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COPHRIGHT DSPOSm 



DIFFERENTIAL AND INTEGRAL 
CALCULUS 



WITH EXAMPLES AND APPLICATIONS 



BY 



GEORGE A. OSBORNE, S.B. 

WALKER PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS 
INSTITUTE OF TECHNOLOGY 



REVISED EDITION 



BOSTON, U.S.A. 
D. C. HEATH & CO., PUBLISHERS 

1906 



a A 



ao2> 



A 



UBRARY of CONGRESS 

Two Copies Received 

JAN 111907 

Cooyriarht Entry 

L. It). Kioto 

iSS o^ XXc„ No. 

COPY B. 



Copyright, 1891 and 1906, 
By GEORGE A. OSBORNE 



PREFACE 

In the original work, tile author endeavored to prepare a text- 
book on the Calculus, based on the method of limits, that should 
be within the capacity of students of average mathematical ability 
and yet contain all that is essential to a working knowledge of the 
subject. 

In the revision of the book the same object has been kept in view. 
Most of the text has been rewritten, the demonstrations have 
been carefully revised, and, for the most part, new examples have 
been substituted for the old. There has been some rearrangement 
of subjects in a more natural order. 

In the Differential Calculus, illustrations of the " derivative" 
aave been introduced in Chapter II., and applications of differentia- 
"ion will be found, also, among the examples in the chapter imme- 
diately following. 

Chapter VII.. on Series, is entirely new. In the Integral Calculus, 
immediately after the integration of standard forms, Chapter XXI. 
has been added, containing simple applications of integration. 

In both the Differential and Integral Calculus, examples illustrat- 
ing applications to Mechanics and Physics will be found, especially 
in Chapter X. of the Differential Calculus, on Maxima and Minima, 
and in Chapter XXXII. of the Integral Calculus. The latter chap- 
ter has been prepared by my colleague, Assistant Professor N. It. 
George, Jr. 

The author also acknowledges his special obligation to his col- 
leagues, Professor H. W. Tyler and Professor F. S. Woods, for 
important suggestions and criticisms. 



CONTENTS 



DIFFERENTIAL CALCULUS 



CHAPTER I 
Functions 

AF.T*. 

I. Variables and Constants 
-7, 9. Definition and Classification of Functions 
8. Notation of Functions. Examples 



PAGES 

1 
1-5 

5-7 



CHAPTER II 



Limit. Increment. Derivative 



10. Definition of Limit 8 

11. Notation of Limit 8 

12. Special Limits (arcs and chords, the base e) . 8-10 
13-15. Increment, Derivative. Expression for Derivative . . .11, 12 

16. Illustration of Derivative. Examples 13-15 

17-21. Three Meanings of Derivative 10-21 

22. Continuous Functions. Discontinuous Functions. Examples . 22-25 



CHAPTER III 
Differentiation 

Algebraic Functions. Examples .... 
Logarithmic and Exponential Functions. Examples 
Trigonometric Functions. Examples . 
Inverse Trigonometric Functions. Examples 
Relations between Certain Derivatives. Examples 



26-39 
39-45 
45-51 
51-57 
57-60 



CHAPTER IV 
Successive Differentiation 



57, 58. Definition and Notation 

60. The nth Derivative. Examples 
60. Leibnitz's Theorem. Examples 



61 
63-65 
65-67 



VI 



CONTENTS 



CHAPTER V 
Differentials. Infinitesimals 



61-63. Definitions of Differential 

64. Formulae for Differentials. 

65. Infinitesimals 



Examples . 



PAGES 

68-70 
71-73 
73,74 



CHAPTER VI 
Implicit Functions 
66. Differentiation of Implicit Functions. Examples 



75-77 



CHAPTER VII 

Series. Power Series 

67,68. Convergent and Divergent Series. Positive and Negative 

Terms. Absolute and Conditional Convergence . . 78, 79 

69-71. Tests for Convergency. Examples 79-85 

72, 73. Power Series. Convergence of Power Series. Examples . 85-87 



CHAPTER VIII 



Expansion of Functions 



74-78. Maclaurin's Theorem. Examples 

79. Huyghens's Approximate Length of Arc 

80,81. Computation by Series, by Logarithms 

82. Computation of -w . 

83-87. Taylor's Theorem. Examples 

89. Rolle's Theorem . 

90-93. Mean Value Theorem . 

94. Remainder .... 



88-93 

93 

94-96 

96,97 

97-100 

101 

101-104 

105 



CHAPTER IX 

Indeterminate Forms 



95. Value of Fraction as Limit 




96, 97. Evaluation of 







Examples 



98-100. Evaluation of g, 0- oo, oo- ex), 0°, 1", oo°. Examples 



106 
106-110 

110-113 



CONTESTS 



vn 



CHAPTER X 

Maxima and Minima of Functions of One Independent 
Variable 

AUTS. PAGES 

101. Definition of Maximum and Minimum Values . . . 114 

102-104. Conditions for Maxima and Minima, Examples . . . 114-119 



105. When 



d.r 



Examples 119-121 



100. Maxima and Minima by Taylor's Theorem. Problems 



121-129 



CHAPTER XI 
Partial Differentiation 

107. Functions of Two or More Independent Variables 

108. Partial Differentiation. Examples .... 

109. Geometrical Illustration 

110. Equation of Tangent Planes. Angle with Coordinate Planes 

Examples 

Ill, 112. Partial Derivatives of Higher Orders. Order of Differentia 

tion. Examples 

113. Total Derivative. Total Differential. Examples 
114-116. Differentiation of Implicit Functions. Taylor's Theorem 

Examples 



130. 131 

131. 132 
133 

133-136 

136-139 
140-144 

144-147 



CHAPTER XII 

Change of the Variables in Derivatives 

117. Change Independent Variable x to y 148, 149 

118. Change Dependent Variable 149 

119. Change Independent Variable z to z. Examples . . 150-152 
120,121. Transformation of Partial Derivatives from Rectangular to 

Polar Coordinates 152-154 



CHAPTER XIII 

Maxima and Minima of Functions of Two or More 
Variables 

122,123. Definition. Conditions for Maxima and Minima 
124. Functions of Three Independent Variables . 



155, 156 
156-161 



CHAPTER XIV 
Curves for Reference 
120-127. Cirsoid. "Witch. Folium of Descartes 



. 162, 163 



Vlll 



CONTENTS 



ARTS. PAGES 

128-130. Catenary. Parabola referred to Tangents. Cubical Pa- 
rabola. Semicubical Parabola 164, 165 

131-134. Epicycloid. Hypocycloid. (-)*+{?)* =1 - a*2/ 2 =« 2 ^-« 6 166,167 

135-145. Polar Coordinates. Circle. Spiral of Archimedes. Hyper- 
bolic Spiral. Logarithmic Spiral. Parabola. Cardioid. 
Equilateral Hyperbola. Lemniscate. Four-leaved Rose. 

r = asm*- 167-172 

o 

CHAPTER XV 

Direction or Curves. Tangents and Normals 

146. Subtangent. Subnormal. Intercepts of Tangent . . 173 

147. Angle of Intersection of Two Curves. Examples . . 174-176 

148. Equations of Tangent and Normal. Examples . . . 176-179 

149-151. Asymptotes. Examples 179-182 

152, 153. Direction of Curve. Polar Coordinates. Polar Subtangent 

and Subnormal 182, 183 

154. Angle of Intersection, Polar Coordinates. Examples . . 183-186 
155, 156. Derivative of an Arc 186-188 

CHAPTER XVI 
Direction op Curvature. Points of Inflexion 

157. Concave Upwards or Downwards 189 

158. Point of Inflexion. Examples 190-192 

CHAPTER XVII 

Curvature. Radius of Curvature. Evolute and 
Involute 

157-161. Curvature, Uniform, Variable 193, 194 

162-164. Circle of Curvature. Radius of Curvature, Rectangular Co- 
ordinates, Polar Coordinates. Examples . . . 195-200 
165. Coordinates of Centre of Curvature ..... 200 

166, 167. Evolute and Involute 201, 202 

168-170. Properties of Involute aud Evolute. Examples . . . 202-205 

CHAPTER XVIII 

Order of Contact. Osculating Circle 

171,172. Order of Contact 206-208 

173. Osculating Curves 208,209 



CONTENTS 



IX 



174. 
175. 



170. 



PAGES 

Order of Contact at Exceptional Points . 209 
To find the Coordinate of Centre, and Radius, of the Oscu- 
lating Circle at Any Point of the Curve .... 209-211 
Osculating Circle at Maximum or Minimum Points. Ex- 
amples 211-213 



CHAPTER XIX 
Envelopes 



177. Series of Curves 

178, 179. Definition of Envelope. Envelope is Tangent 

180-182. Equation of Envelope 

183, Evolute of a Curve is the Envelope of its Normals. 



214 

. 214,215 

. 215-217 

Examples 217-221 



INTEGRAL CALCULUS 

CHAPTER XX 
Integration of Standard Forms 

184, 185. Definition of Integration. Elementary Principles . . 223-225 
186-190. Fundamental Integrals. Derivation of Formulae. Examples 225-240 



CHAPTER XXI 

Simple Applications of Integration. Constant of 
Integration 

191,192. Derivative of Area. Area of Curve. Examples . 

195. Illustrations. Examples 



241-244 

244-248 



CHAPTER XXII 
Integration of Rational Fractions 

194, 195. Formulae for Integration of Rational Functions 
Operations 

196. Partial Fractions . 

197. Case I. Examples 

198. Case II. Examples 

199. Case III. Examples 

200. Case IV. Examples 



p 


reliminary 






249 








250 








250-253 








254-256 








256-259 






» . 


260-262 



CONTENTS 



CHAPTER XXIII 
Integration of Irrational Functions 



202. Integration by Rationalization 

p 
203, 204. Integrals containing (ax + &)«, 
206, 207. Integrals containing V± x 2 + q 

208. Integrable Cases 



(ax + b) s . Examples 
z + b. Examples 



PAGES 

263 

263-266 
266-268 



CHAPTER XXIV 

Trigonometric Forms readily Integrable 

209-211. Trigonometric Function and its Differential. Examples . 270-272 

212, 213. Integration of tan n x dx, cot" x dx, sec n xdx, cosec n xdx . 273, 271 

214. Integration of tan m x sec n x dx, cot m x co&ec n xdx. Examples 274-276 

215. Integration of sin™ x cos w x dx by Multiple Angles. Examples 276-278 

CHAPTER XXV 
Integration by Parts. Reduction Formulae 

216. Integration by Parts. Examples 279-282 

217. Integration of e ax sin nx dx, e ax cos nx dx. Examples . . 283,284 
218-222. Reduction Formulae for Binomial Algebraic Integrals. Deri- 
vation of Formulae. Examples 284-291 

223, 224. Trigonometric Reduction Formulae. Examples . . . 291-294 



CHAPTER XXVI 

Integration by Substitution 

p 
226. Integrals of f(x 2 )xdx, containing (a -f bx 2 )i. Examples 



227. Integrals containing Vcfi 



y/x 2 ± a 2 by Trigonometric 



Substitution. Examples 

228-232. Integration of Trigonometric Forms by Algebraic Substitu- 
tion. Examples 

233. Miscellaneous Substitutions. Examples . 



295, 296 
296-299 

299-304 
304, 305 



CHAPTER XXVII 

Integration as a Summation. Definite Integrals 

234. Integral the Limit of a Sum 

235-237. Area of Curve. Definite Integral. Evolution of Definite 
Integral , 



306 



306-309 



CONTEXTS 



XI 



ART?. PAGES 

288, 239. "Definition of Definite Integral. Constant of Integration. 

Examples 310-314 

240-242. Sign of Definite Integral. Infinite Limits. Infinite Values 

of/(.r) 314-317 

243-245. Change of Limits. Definite Integral as a Sum . . . 317-319 

CHAPTER XXVIII 

Application- of Integration to Plane Curves. 
Application to Certain Volumes 

246. 247. Areas of Curves, Rectangular Coordinates. Examples . 320-324 

24S. Areas of Curves. Polar Coordinates. Examples . . . 325-327 

249. Lengths of Curves. Rectangular Coordinates. Examples . 327-330 

250. Lengths of Curves, Polar Coordinates. Examples . . 330-332 

251. Volumes of Revolution. Examples 333-335 

252, 253. Derivative of Area of Surface of Revolution. Areas of Sur- 
faces of Revolution. Examples 336-339 

254. Volumes by Area of Section. Examples .... 340-342 



255-25' 



CHAPTER XXIX 
Successive Integration 

Definite Double Integral. Variable Limits. Triple Integrals. 

Examples . 343-345 



CHAPTER XXX 
Applications of Double Integration 

258-262. Moment of Inertia. Double Integration, Rectangular Co- 
ordinates. Variable Limits. Plane Area as a Double 
Integral. Examples 

26:5-265. Double Integration, Polar Coordinates. Moment of Inertia 

Variable Limits. Examples 

266. Volumes and Surfaces of Revolution, Polar Coordinates 
Examples 



346-350 

350-353 



353, 354 



CHAPTER XXXI 

Surface, Volume, and Moment of Inertia of Ant Solid 

267. To find the Area of Any Surface, whose Equation is given 
between Three Rectangular Coordinates, x, y, z. Ex- 
amples 355-360 



Xll 



CONTENTS 



268. To find the Volume of Any Solid bounded by a Surface, 

whose Equation is given between Three Rectangular Co- 
ordinates, jc, y, z. Examples 361-363 

269. Moment of Inertia of Any Solid. Examples . . . 363, 364 



CHAPTER XXXII 

Centre of Gravity. Pressure of Eldids. 
Eorce of Attraction 

270,271. Centre of Gravity. Examples 

272, 273. Theorems of Pappus. Examples . 

274. Pressure of Liquids. Examples . 

275. Centre of Pressure. Examples 

276. Attraction at a Point. Examples 



365-369 
369, 370 
370-373 
373-375 

375-377 



CHAPTER XXXIII 
277. Integrals for Reference .... 

Index . . 



378-385 

386-388 



DIFFERENTIAL CALCULUS 

CHAPTER I 
FUNCTIONS 

1. Variables and Constants. A quantity which may assume an 
unlimited number of values is called a variable. 

A quantity whose value is unchanged is called a constant. 
For example, in the equation of the circle 

x 2 +y° = a 2 , 

x and y are variables, but a is a constant. For as the point whose 
coordinates are x, y, moves along the curve, the values of x and y 
are continually changing, while the value of the radius a remains 
unchanged. 

Constants are usually denoted by the first letters of the alphabet, 
a, b, C, a, (3, y, etc. 

Variables are usually denoted by the last letters of the alphabet, 

*, y, z, <t>, «A, etc - 

2. Function. When one variable quantity so depends upon an- 
other that the value of the latter determines that of the former, the 
former is said to be a function of the latter. 

For example, the area of a square is a function of its side ; the 
volume of a srjhere is a function of its radius ; the sine, cosine, and 
tangent are functions of the angle ; the expressions 



x 2 , log (V 9 + 1), V*(* + l), 
are functions of x. 



2 DIFFERENTIAL CALCULUS 

A quantity may be a function of two or more variables. For 
example, the area of a rectangle is a function of two adjacent sides; 
either side of a right triangle is a function of the two other sides ; 
the volume of a rectangular parallelopiped is a function of its three 
dimensions. 

The expressions 

x 2 + xy + y 2 , \ g(x 2 + y 2 ), a x+ ^, 

are functions of x and y. 

The expressions 

xy + yz + zx, ^^~~f log(x 2 + y~z), 
are functions of x, y, and z. 

3. Dependent and Independent Variables. If y is a function of x, 
as in the equations 



y = x 2 , y = tan 4 a?, y = e x -f- 1 



x is called the independent variable, and y the dependent variable. 

It is evident that when y is a function of x, x may be also regarded 
as a function of y, and the positions of dependent and independent 
variables reversed. Thus, from the preceding equations, 

x=Vy, x = ±tan- 1 y, x = log e (y-l). 

In equations involving more than two variables, as 

z + x — y = 0, iv + wz + zx + y = 0, 

one must be regarded as the dependent variable, and the others as 
independent variables. 

4. Algebraic and Transcendental Functions. An algebraic function 
is one that involves only a finite number of the operations of addi- 
tion, subtraction, multiplication, division, involution and evolution 
with constant exponents.* All other functions are called transcen- 
dental functions. Included in this class are exponential, logarithmic, 
trigonometric or circular, and inverse trigonometric, functions. 

Note. — The term "hyperbolic functions" is applied to certain 
forms of exponential functions. See page 00. 

* A more general definition of Algebraic Function is, a function whose rela- 
tion to the variable is expressed by an algebraic equation. 



FUNCTIONS 3 

5. Rational Functions. A polynomial involving only positive 
integral powers of x, is called an integral function of x\ as, for 
example, 2 + x - 4 .r + 3 x\ 

A. rational fraction is a fraction whose numerator and denominator 
are integral functions of the variable ; as, for example, 

a, r 3 + 2q;-l 
x* + x*-2x' 

A rational function of x is an algebraic function involving no frac- 
tional powers of x or of any function of x. 

The most general form of such a function is the sum of an integral 
function and a rational fraction ; as, for example, 

3x 2 -2x 



2 ,r rf- x - 1 + 



x- + l 



6. Explicit and Implicit Functions. When one quantity is ex- 
pressed directly in terms of another, the former is said to be an 
explicit function of the latter. 

For example, y is an explicit function of x in the equations 

y = a ? + 2x, y = Va 2 + 1. 

"When the relation between y and x is given by an equation con- 
taining these quantities, but not solved with reference to y, y is said 
to be an implicit function of x, as in the equations 

axy -f bx 4- cy + d = 3 y -\- log y = x. 

Sometimes, as in the first of these equations, we can solve the 
equation with reference to y, and thus change the function from 
implicit to explicit. Thus we find from this equation, 

_bx±d m 

ax + c 

7. Single-valued and Many-valued Functions. In the equation 

y = X- - 2 .r, 

for every value of x, there is one and only one value of y. 
Expressing x in terms of y, we have 

x = 1 ± Vy + 1- 



4 DIFFERENTIAL CALCULUS 

Here each value of y determines two values of x. In the former 
case, y is a single-valued function of x. 

In the latter case, x is a two-valued function of ?/. 

An w-valued function of a variable x is a function that has n 
values corresponding to each value of x. 

The inverse trigonometric function, tan -1 x, has an unlimited num- 
ber of values for each value of x. 

8. Notation of Functions. The symbols F(x),f(x), <£(#), \f/(x), 
and the like, are used to denote functions of x. Thus instead of " y 
is a function of x," we may write 

y=f(x), or y = <f>(x). 

A functional symbol occurring more than once in the same prob- 
lem or discussion is understood to denote the same function or 
operation, although applied to different quantities. Thus if 

• f(x) = x* + 5, (1) 

then f(y)=y* + &, /( ft ) = a 2 + 5, 

/(a + l) = (a + l) 2 + 5 = a* + 2a + 6, 

/(2) = 2 2 + 5 = 9, /(1) = 6. 

In all these expressions /( ) denotes the same operation as de- 
fined by (1) ; that is, the operation of squaring the quantity and 
adding 5 to the result. 

Functions of two or more variables are expressed by commas be- 
tween the variables. 

Thus if / (x, y) = x 2 + 3 xy - f, 

then f(a, b) = a 2 + 3 ab - b\ f(b, a) = b 2 + 3 ba - a 2 . 

jf(3, 2) = 3 2 + 3-3-2 - 2 2 = 23. f(a, 0) = a 2 . 

If 4>(x, y, z) = x s + yz-y 2 + 2, 
then <f>(3, l,-l) = 3 3 + l(-l)-l 2 + 2 = 27; 
*(a, 0, 0) = a 3 - b 2 + 2 ; <K0, 0, 0) = 2. 



FUNCTIONS 5 

9. Inverse Function. If y is a given function of x, represented by 

y = +(*), (i) 

and if from this relation we express x in terms of y, so that 

■'■ = •/<(.'/), (2) 

then each of the functions <f> and \p is said to be the inverse of the 
other. 
. For example, if V = * = <£<»> 

then x =VS = ^O/)- 

Here \p, the cube root function, is the inverse of cf>, the cube 
function. 

If V = a* = <Kx), 

then x = log a y = xp(y). 

Here «/r, the logarithmic function, is the inverse of <f>, the exponential 

function. n 

x -4- 2i 

Again, suppose y =-, -=<f>(x) (3) 

r — x 

v — 2 
From this we derive x = ' j = \p(y) (4) 

Here xp as defined by (4) is the inverse of <f> as defined by (3). 
The notation ^> _1 is often employed for the inverse function of <j>. 

Thus, if y = <£(#), x = <£ _1 0/). 

if y=f(?), x=f~ 1 (y). 

The student is already familiar with this notation for the inverse 
trigonometric functions. 

If y = sin x, x = sin -1 ?/. 



EXAM PLES 
1. Given 2x* — 2 xy + y 2 = a 2 ; 

change y from an implicit to an explicit function. 
y = x ± Va 2 — x 2 . 



6 DIFFERENTIAL CALCULUS 

ft 2. Given sin (a; — y)~ m sin ?/ ; 

change y from an implicit to an explicit function, 

Ans. y = tan' 1 — 

m + cos x 

^3. Given /(a?) = 2^-3^+^+2 ; 

find /(l), /(2), /(|), /(- 1), /(0). 

Show that / (a + 1) -/(*) = 6 a; 2 , 

fix + ft) =/(*) + (6 a; 2 - 6 i» + l)h + (6x- 3)/i 2 + 2 fc s . 

H. Given F(x) = (x 2 -1) 2 ; 

show that i^(a? + 1) - F(x — 1) = 8 X s . 

h. Griven.f(x) = ^t^l: } find/(0), /(*)+/(-«). 

Show that / (2a?) -/(- 2 ag = [/ (a:)] 2 - [/ (- a?)] 2 . 

%K ^. If <£ (0) = c», <£ (a -\-b) = cj>(a)cj> (6). 

Show that the same relation holds for the function 
$(0) = cos 4- V"^ sin (9, 
giving if/(a + b) = if, (a) ^(6). 

" 7 - If >«-sfey 

show that the inverse function is of the same form. 

8. If <f>(x) = X , find the inverse function of <j>. 
ax— c 

Compare the two functions when b = c. 

V9. If /(«) = log.(«+'V?=l), 

show that /- 1 (a;) = a * + *y • 

10. If / (a^) = ax* + 2 toy + q/ 2 ; find /(l, 2), / (y, - a?). 

Show that 
/ (a? + ft, y + ft) =/(«, 2/.) + 2(aa? + by)h + 2(bx + cy) ft +f(h, ft). 



11. Given 



FUNCTIONS 

\m-\-n 



«£(m,n) 



\m |w 



where m, », are positive integers; show that 
<f> (m, w + 1) + <£ (m + 1, ») = <K m + 1? ?i + !)• 



12. Given 



/ fe & ») 



«, ?/, 


z 


2, X, 


y 


y» *> 


X 



show that f{y + z,z + x, x + y) = 2/ (a;, y, 0). 



CHAPTER II 
LIMIT. INCREMENT. DERIVATIVE 

10. Limit. When the successive values of a variable x approach 
nearer and nearer a fixed value a, in such a way that the difference 
x — a becomes and remains as small as we please, the value a is 
called the limit of the variable x. 

The student is supposed to be already somewhat familiar with the 
meaning of this term, of which the following illustrations may be 
mentioned. 

The limit of the value of the recurring decimal .3333 ..., as the 
number of decimal places is indefinitely increased, is ^. 

The limit of the sum of the series 1 + -J- + -J- + -J-H — , as the number 
of terms is indefinitely increased, is 2. 

The limit of the fraction - ~ a , as x approaches a, is 3 a 2 . 

x — a 

The circle is the limit of a regular polygon, as the number of sides 
is indefinitely increased. 

The limit of the fraction — — , as 6 approaches zero, is 1, provided 

6 
6 is expressed in circular measure. 

11. Notation of Limit. The following notation will be used : 
"Lim^" denotes "The limit, as x approaches a, of." 

x 2 —a 2 
For example, Lim z=a — == 2. 

xr — ax 

Lim A=0 (2 x 2 - hx + h 2 ) =2x>. 

12. Some Special Limits. There are two important limits required 
in the following chapter. 

(a) Lim 0=o , 6 being in circular measure. 



LIMIT 



Let the angle ADA' = 2 0. and let a be the radius of the arc AC A!. 

From geometry, ABA' < AC A' ; 
sin 



that is, 2 a sin < 2 aO, ^- v 





(i) 



Also from geometry, AC A' < ADA ; 



that is, 2a0 <2 a tan 0, 



sin 




sin fl 
cos 



>e, 



> cos 0. 



(2) 



Hence by ^1) and (2), is inter- 

u 
mediate in value between 1 and cos 0. 
As approaches zero, cos approaches 1. 

sin 9 .. 




Hence 



Lini„ =0 



The student will do well to compare the corresponding values of 
and sin 6, taken from the tables, for angles of 5°, 1°, and 10'. 



Angle 





sin 6 


5° 


— = .0872605 
36 




.0871557 


1° 


-5- = .0174533 
180 




.0174524 


10' 


'■ =.0029089 
1080 




.0029089 


(b) Lim 2=ae M + 


IV 

— ] Before deriving this limit 

z )' 


let us compute 


the value of the. ex 


pression for increasing values of 

(1 + 1) 2 = 2.25 

(1 + 1) 5 = 2.48832 


z. Thus, 




(1 + to) 10 = 2.59374 . 


• 




(1.01) 100 = 2.70481 






(1.001) 1000 = 2.71692 






(1.0001) 10000 = 2.71815 






(1.00001) 100000 = 2.71827 






(1.000001) 1000000 = 2.71& 


18 





10 DIFFERENTIAL CALCULUS 

The required limit will be found to agree to five decimals with the 
last number, 2.71828. 

By the Binomial Theorem, 
which may be written ^ / 1\ / 2 

+ iy = i + i + izl + v 1 ~w~~z/ 

zj , [2 |3 

When z increases, the fractions -, - etc., approach zero, and we 

T z z 

have 

This quantity is usually denoted by e, so that 



e = 


1+7 + %+£ + £ + '—- 




1 |2 [3 |4 


The value of e can be easily calculated to any desired number of 


decimals by computing the values of the successive terms of this 


series. For seven decimal pi 


ices the calculation is as follows: 


2) 


1. 
1. 


3) 


.5 


4) 


.166666667 


5) 


.041666667 


6) 


.008333333 


7) 


.001388889 


8) 


•000198413 


9) 


.000024802 


10) 


.000002756 


11) 


.000000276 




.000000025 



e= 2.7182818--. 
This quantity e is the base of the Napierian logarithms. 

* For a rigorous derivation of this limit, the student is referred to more ex- 
tensive treatises on the Differential Calculus. 



DERIVATIVE 



11 



13. Increments. An increment of a variable quantity is any addi- 
tion to its value, and is denoted by the symbol A written before this 
quantity. Thus Ax denotes an increment of x, Ay, an increment of y. 

For example, if we have given 

y = A 

and assume x = 10, then if we increase the value of x by 2, the value 
of y is increased from 100 to 141, that is, by 41. 

In other words, if we assume the increment of x to be A& = 2, we 
shall find the increment of y to be A?/ = 44. 

If an increment is negative, there is a decrease in value. 

For example, calling x = 10 as before, in y = x 2 , 

if Ax = - 2, then Ay = - 36. 

14. Derivative. With the same equation, 

y = x 2 , 
and the same initial value of x, 

z = 10, 

let us calculate the values of Ay corresponding to different values of 
A.'-. We thus find results as in the following table. 



If Ax = 


then Ay = 


and -^- = 

Ax 


3. 


09. 


23. 


2. 


44. 


22. 


1. 


21. 


21. 


0.1 


2.01 


20.1 


0.01 


0.2001 


20.01 


0.001 


0.020001 


20.001 


h 


20 h + 1C- 


20+ ft 



The third column gives the value of the ratio between the incre- 
ments of ./■ and of y. 

It appears from the table that, as Ax diminishes and approaches 
zero. Ay also diminishes and approaches zero. 



12 DIFFERENTIAL CALCULUS 

The ratio — ^ diminishes, but instead of approaching zero, ap- 
proaches 20 as its limit. 

This limit of — ^ is called the derivative of y with respect to x, 

and is denoted by -^. In this case, when x = 10, the derivative 

^ = 20. ** 

dx 

It will be noticed that the value 20 depends partly on the func- 
tion y = x 2 , and partly on the initial value 10 assigned to x. 

Without restricting ourselves to any one initial value, we may ob- 
tain -^ from y = x 2 . 
dx 

Increase x by Ax. Then the new value of y will be 
y' = (x + Ax) 2 ; 
therefore, Ay = y 1 — y = (x + Ax) 2 — x 2 = 2xAx + (Ax) 2 . 

Dividing by Ax, Ay = 2 x + ^ 

Ax 

The limit of this, when Ax approaches zero, is 2x. 

Hence, — = 2 x. 

dx 

The derivative of a function may then be defined as the limiting 
value of the ratio of the increment of the function to the increment of the 
variable, as the latter increment approaches zero. 

It is to be noticed that -2 is not here defined as & fraction, but as 
dx A 

a single symbol denoting the limit of the fraction — ^. The student 

will find as he advances that — has many of the properties of an 
ordinary fraction. 

The derivative is sometimes called the differential coefficient. 

15. General Expression for Derivative. In general, let 

y -/(»)■ 

Increase x by Ax, and we have the new value of y } 

y'=f(x-\-Ax). 



DERIVATIVE 



13 



Ay = y' - y =/(* + A ^) — /(*)i 

A?/ = /(* + A.r)-/(.r) 
A.y A# 

(f.i' A# 



Geometrical Illustration. The process of finding the derivative 
from y = x-, may be illustrated by a squared ■ 

Let x be the length of the side OP, and y the area of the square 
on OP. 

That is. y is the number of square units 
corresponding to the linear unit of x. 

When the side is increased by PP', the 

area is increased by the space between the 

squares. 

Ay_ 



That is, Ay=2xAx+(Ax)* =^=2x+Ax, 

Ax 



dy T • A?/ 

ax Ax 



2x. 



Ay 


y 


X 




X 




Ax 



p P' 



16. From the definition of the derivative we have the following 
process for obtaining it : 

(a) Increase x by Ax, and by substituting x -f- Ax for x, deter- 
mine y + Ay, the new value of y. 

(b) Find Ay by subtracting the initial value of y from the new 
value. 

(c) Divide by Ax, giving — • 

Ax 

(d) Determine the limit of _ _?, as Ax approaches zero. This 

limit is the derivative — . 
dx 

Apply this process to the following examples. 

EXAMPLES 
1 . y = 2 .r 3 — 6 x + 5. 

Increasing x by Ax, we have 

y + Ay = 2(x + Axf-6(x + Ax) + 5; 

therefore, Ay = 2 (x + Ax) 3 — 6 (x + Ax) + 5 — 2 x' + 6x—5 

= (6 x 2 - 6) Ax + 6 x(Ax) 2 + 2(A x) 3 . 



14 DIFFERENTIAL CALCULUS 

Dividing by Ax, 

-^ = 6x 2 -6 + 6xAx + 2(Ax) 2 . 

Ax v } 

* = Lim A ^/ = 6x 2 -6. 
dx Ax 

2. y = 



y + Ay = 
Ay = 



x + 1 

x + Ax 
x + Ax + 1 ' 

x + Ax x Ax 



x + Ax + 1 z + 1 (x-f Ax-|-l)(x + l) 

Ay = 1 

Ax (x + Ax-fl)(x + l)' 

^-Lira ^ = _J: 

dx~ ^°Ax (x + 1) 2 

3. 2/= Vx. 

y -\- Ay = Vx -f Ax, 



A?/ = Vx + Ax — Vx, 



Ay _ Vx -f- Ax — V# _ 
Ax Ax 

The limit of this takes the indeterminate form -. But by 

J 

rationalizing the numerator, we have 

Ay _ Ax 1 



Ax Ax(Vx + Ax+Vx) Vx+Ax-fVx 



dx Az=0 Ax 2V 

4. y = x 4 - 



^4. 2/ = aj 4 -2x 2 + 3x-4, ^ = 4x 3 -4x + 3. 

dx 



5. ?/=(x-a) 3 , ^/ = 3(x-a) 2 . 

dx 



DERIVATIVE 15 

6. y=(« + 2)(3-2«) J g = -4*-l. 



mo; 

11 = 



x = 



> 
it — X 

2.V-T. 


» + »* 

.r + a 2 




(*-!)" 



% 


mn 


da 


\n-xf 


dx 


9 


cty 


"<* + *>* 


dx 


(«j + a) 


dx _ 
dt ~ 


« + l 


dy _ 


1 



10. .-/ = 

V ,= 

*12. ^/=V^T2 



8 

13. y = x\ 



14. 2/=V« 2 -^, 



15. s = ± 



16. Show that the derivative of the area of a 'circle, with respect 
to its radius, is its circumference. 

17. Show that the derivative of the volume of a sphere, with, 
respect to its radius, is the surface of the sphere. 

"We shall now give some illustrations of the meaning of the deriva- 
tive. 



dx 


2Vz-r-2 


dy = 
dx 


Zx% 

2 ' 


dy _ 


X 


dx 


-Vcr—x 2 


dx _ 
dt~ 


1 
2$ 



16 



DIFFERENTIAL CALCULUS 




17. Direction of a Plane Curve. This is one of the simplest and 
most useful interpretations of the derivative. 

Let P be a point in a curve determined by its equation y = f(x), 
and PT the tangent at P. 

Let OM=x, MP = y. 

If we give x the increment 
Ax = MN, y will have the in- 
crement Ay = PQ. 

Draw PQ. Then 

Now if Ax diminish and 
approach zero, Ay will also 
approach zero, the point Q 
will move along the curve 

towards P, and PQ will approach in direction PT as its limit. 
Taking the limit of each member of (1), we have 

tan TPR = Lim Ax _ ^ = ^ 
Ax dx 

That is, the derivative — , at any point of a curve, is the trigono 
dx 

metric tangent of the in- 
clination to OX of the 
tangent line at that 
point. 

This quantity is de- 
noted by the term slope. 

The slope of a straight 
line is the tangent of its 
inclination to the axis 
of X 

The slope of a plane 
curve at any point is the 
slope of its tangent at 
that point. 

Thus, -^, at any point of a curve, is the slope of the curve at that 
point. 




DERIVATIVE 17 

x 2 
For example, consider the parabola x 2 = ±py, y = - — 

The slope of the curve is — = — • 
1 dx 2p 

At 0, where x = 0, the slope = 0, the direction being horizontal. 

At L, where x = 2 p, the slope = 1, corresponding to an inclination 
of 45° to the axis of X 

Beyond L the slope increases towards oo, the inclination increasing 
towards the limit 90°. 

For all points on the left of Y", x is negative, and hence the slope 
is negative, the corresponding inclinations to the axis of X being 
negative. 



18. Velocity in Terms of a Variable t denoting Time. A body moves 
over the distance OP = s in the time t, s being a function of t; it is 
required to express the velocity at the point P. 

Let As denote 

the distance ir £ — - — ±-, 

PP traversed 

in the interval At. If the velocity were uniform during this interval, 

As 

it would be equal to 

As 
For a variable velocity, — is the average or mean velocity between 

P and P, and is more nearly equal to the velocity at P the less we 
make At. 

That is, the velocity at P — Lim Afe0 -r. — -r.' 



At dt 



ds 

If v denote this velocity, v = — • 

Thus, — is the rate of increase of s. 
dt 



Similarly, -vr and -rr are the rates of increase of x and y respectively. 



18 



DIFFERENTIAL CALCULUS 



Then the velocity, 
and the acceleration, 



v = 



3? 



6t. 



19. Acceleration. The rate of increase of the velocity v is called 
acceleration. 

If we denote this by a, we have by the preceding article, 

dv 
a — — 
dt 

For example, suppose a body moves so that 

s = t\ 

ds _ 

dt' 

dv 

dt' 

20. Rates of Increase of Variables. For further illustrations of the 
derivative, consider the two following problems : 

Problem 1. A man 
walks across the street 
from A to B at a uniform 
rate of 5 feet per second. 
A lamp at L throws his 
shadow upon the wall 
MN. AB is 36 feet, and 
BL 4 feet. How fast is 
the shadow moving when 
he is 16 feet from A? 
When 26 feet ? When 
30 feet? 

Let P and Q be si- 
multaneous positions of man and shadow. 
y _ BL 4 4 x 

x~PB 




Let AP = x, AQ = y. 



Then 



V 



(1) 



36 — x 36 —a; 

When he walks from P to P', the shadow moves from Q to Q'. 
That is, when Ax=PP', \y=QQ'. 

Let At be the interval of time corresponding to Ax and Ay. 

Ay = At^ 
Ax Ax ' 
At 



Then we may write 



(2) 



DERIVATIVE 



19 



If now we suppose At to diminish indefinitely, Ay and A.i* will 
also diminish indefinitely, and we have for the limits of the two 
members of ^2), f? 

dy dt rate of increase of ?/ 
dx — dx rate of increase of x 
dt 



Art. 18. 



That is, 



velocity of shadow at any point Q _ dy 
velocity of man dx 



Finding the derivative of (1), we have 

dy = 144 
dx (36 - xj 
Hence, 
velocity of shadow at any point Q = 

144 



144 



(36 - x) 
720 



(36 — x} 2 
(5 feet per second) 



See Ex. 8, Art. 16. 
(velocity of man) 



- feet per second 
(oo — X)- 

= 1.8 feet per second, when x = 16 ; 

= 7.2 feet per second, when x = 26 ; 

= 20 feet per second, when x = 30. 

Problem 2. The top of a ladder 20 feet long rests against a wall. 
The foot of the ladder is moved away from the wall at a uniform 
rate of 2 feet per second. 
How fast is the top moving, 
when the foot is 12 feet 
from the wall ? AVhen 16 
feet from the wall ? 

Suppose PQ to be one 
position of the ladder. 

Let 

AP= x, AQ = y. 

Then 

y = V400 - x 2 . (3) 




20 



DIFFERENTIAL CALCULUS 



When the foot moves from Pto P', the top moves from Q to Q\ 
That is, if Ax = PP', Ay = QQ'. 
In the same way as in Problem 1, 

Ay 

Ay _ At 

Ax ~ Ax 
At~ 



And from this, 



that is, 



From (3), 



dy 
dy _dt 
dx~ dx' 

dt 

velocity of top at Q _ dy 
velocity of foot dx 

dy _ — x 
<^~V400-;/ 



See Ex. 14, Art. 16. 



Hence, 
velocity of top at any point Q = 



V400-X 2 
2x 



V400-^ 2 



= (velocity of foot) 
feet per second. 



The negative sign is explained by noticing from the figure that y 
decreases when x increases. Hence the rates of increase of x and y 
have different signs. 

When x — 12, velocity of top = — 11 feet per second. 

When x = 16, velocity of top = — 2| feet per second. 

Aw 
From these problems it appears that, while -^ is the ratio between 

dv 
the increments of y and x, —■ is the ratio betiveen the rates of increase 
* ' dx J 

of these variables. 



DERIVATIVE 



21 



21. Increasing and Decreasing Functions. If the derivative of a 
junction of x is positive, the junction increases when x increases; and if 

the derivative is negative, the function decreases ivhen x increases. 

For if the derivative — , which is the ratio between the rates of 
dx 

increase of the variables (see conclusion of Art. 20), is positive, it 
follows that these rates must have the same sign ; that is, y increases 
when x increases, and decreases when x decreases. 

But if -^ is negative, the rates must have different signs ; that is, 
ax 

y decreases when x increases, and increases when x decreases. 

This is also evident geometrically by regarding -^ as the slope of 
a curve. 

As we pass from A to B, y increases as x increases, but from B to 
C, y decreases as x in- 
creases. 

Between A and B the 

slope — is positive ; be- 
tween B and C, negative. 

In the former case y is 
said to be an increasing 
function ; in the latter 
case, a decreasing function. 

For example, consider 

the function y = x 3 , from which we find -^ = 3x*. 

rh ^ x 

Since — is positive for all values of x, the function y = x 3 is an 
dx 

increasing function. 

1 




If we take y 



— , we find — 
x dx 



1 



Here we have a decreasing function with a negative derivative. 
Another illustration is Ex. 1, Art. 16, 

y = 2 ar 3 - 6 x + 5, ^ = 6 (a? - 1). 

ax 

When x is numerically less than 1, y is a decreasing function. 
When x is numerically greater than 1, y is an increasing function. 



22 



DIFFERENTIAL CALCULUS 



22. Continuous Function. A function, y=f(x), is said to be 
continuous for a certain value x , of x, when y = f (afc) is a definite 
quantity, and A?/ approaches zero as Ax Q approaches zero, Ax being 
positive or negative. 

The latter condition is sometimes expressed, "when an infinitely 
small increment in x produces an infinitely small increment in y." 

The most common case of discontinuity of the elementary functions 
(algebraic, exponential, logarithmic, trigonometric and inverse trigo- 
nometric, functions) is when the function is infinite. 



Y 


a 


^ 






"-— — - ° 


\ 


A 



For example, consider the function y = 



1 



which is continuous 



for all values of x except x = a. 

When x = a, y = oo, that is, y can be made as great as we please by 
taking x sufficiently near a. Also when x<a, y is negative, and 
when x>a, y is positive. 

There is a break in the curve when x = a, and the function is said 
to be discontinuous for the value x = a. 



DERIVATIVE 



23 




The function y 



is likewise discontinuous when x = a. 



(x-ay 
This function being positive for all values of x, the -two branches 
of the curve are above the axis of x. 

Likewise the functions, tan x, sec x, are discontinuous when x = — • 

Z 

In general, if/(.r) = oo, when x = a, there is a break in the curve 
y=f(x) corresponding to x = a, and both the curve and the function 

are then discontinuous. I 

2 X 4- 2 
Another form of discontinuity is seen in the function y = — , 

when a = 0. 2*+l 

Here y approaches two limits, according as x approaches zero 
through positive or negative values. 



Lim z = + -^— =1. 

2'+l 

We see that when x = the 
curve jumps from y=2 to y=l, 
that is from B to A. 

The function is discontinu- 
ous for x = 0. 

It is to be noticed that the 
definition of the derivative 



Lim . 



i 

r+i 



= 2. 



24 



DIFFERENTIAL CALCULUS 



implies the continuity of the function. For — ^ cannot approach a 

Ax 

limit, unless Ay approaches zero when Ax approaches zero. 

The converse is not true. There are continuous functions which 

have no derivative, but they are never met with in ordinary 

practice. 

EXAMPLES 



1. The equation of a curve is y 



x> + 2. 



(a) Find its inclination to the axis of x, when x = 0, and 

when x = 1. Ans. 0° and 135°. 

(b) Find the 
points where the 
curve is parallel to 
the axis of X. 
Ans. x=0 and x=2. 

(c) Find the 
points where the 
slope is unity. 

Ans. £C = (1±V2). 

(d) Find the 
point where the direc- 
tion is the same as 
that at x = 3. Ans. x = — l. 




> 2. In Problem 1, Art. 20, when will the velocity of the shadow 
be the same as that of the man ? Ans. When AP= 24 ft. 

When one quarter, and when nine times, that of the man ? 

Ans. When AP = 12 ft., and 32 ft. 

^3. A circular metal plate, radius r inches, is expanded by heat, 
the radius being expanded m inches per second. At what rate is 
the area expanded ? Ans. 2 irrm sq. in. per sec. 

4. At what rate is the volume of a sphere increasing under the 
conditions of Ex. 3 ? Ans. 4 Trrra cu. in. per sec. 



DERIVATIVE 25 

5. The radius of a spherical soap bubble is increasing uniformly 
at the rate of y L inch per second. Find the rate at which the 
volume is increasing when the diameter is 3 inches. 

Ans. ^ = 2.827 cu. in. per sec. 

6. In Exs. 5, 7, Art. 16, is y an increasing or a decreasing function ? 
Is " an increasing or a decreasing function of x ? 



» + l 



o 



7. In the Example 1, above, for what values of x is y an increas- 
ing function of x, and for what values a decreasing function ? 

8. Find where the rate of change of the ordinate of the curve 
y = X s — 6.i~ + 3.r + o, is equal to the rate of change of the slope of 
the curve. Ans. x= 5 or 1. 

x 3 

9. When is the fraction — - — - increasing at the same rate as a?? 
x 2 + a 2 

Ans. When x 2 = a 2 . 

10. If a body fall freely from rest in a vacuum, the distance 
through which it falls is approximately s = 16 t 2 , where s is in feet, 
and t in seconds. Find the velocity and acceleration. What is the 
velocity after 1 second ? After 4 seconds ? After 10 seconds ? 

Ans. 32, 128, and 320 ft. per sec. 



CHAPTER III 
DIFFERENTIATION 

23. The process of finding the derivative of a given function is 
called differentiation. The examples in the preceding chapter illus- 
trate the meaning of the derivative, but the elementary method of 
differentiation there used becomes very laborious for any but the 
simplest functions. 

Differentiation is more readily performed by means of certain 
general rules or formulae expressing the derivatives of the standard 
functions. 

In these formulae u and v will denote variable quantities, func- 
tions of x ; and c and n constant quantities. 

It is frequently convenient to write the derivative of a quantity u, 

— u instead of — , 
dx dx 

the symbol — denoting " derivative of ." 
dx 

Thus A^ ~r v) ^ t k e derivative of (u -f- v\ may be written — (u-\- v), 
dx dx 

24. Formulae for Differentiation of Algebraic Functions. 

i. ^=1. 

dx 

II. — = 0. . 
dx 

TTT — ( 4- } — — -l- — . 
dx dx dx 

26 



DIFFERENTIATION 27 



TTT d f N du . dv 

IV. — («r) = v h ?< — • 

dx dx dx 



V d , ^ __ du 
dx dx 



V u 



vi A/ 7 ^— . dx dx 



dx\vj v 2 

VII. — (vr)=nu n ~ 1 —. 

dx dx 

These formulae express the following general rules of differenti- 
ation : 

I. TJie derivative of a variable with respect to itself is unity. 
II. TJie derivative of a constant is zero. 

III. TJie derivative of the sum of two variables is the sum of their 
derivatives. 

IV. TJie derivative of tJie product of two variables is tJie sum of 
the products of eacJi variable by tJie derivative of the other. 

V. TJie derivative of tJie product of a constant and a variable is 
tJie product of the constant and.tJie derivative oftJie variable. 

VI. TJie derivative of a fraction is tJie derivative of tJie numerator 
multiplied by tJie denominator minus tJie derivative of tJie denomi- 
nator multiplied by tJie numerator, this difference being divided by the 
square of 'the denominator. 

VII. TJie derivative of any power of a variable is tJie product of the 
exponent, tJie power witJi exponent diminisJied by 1, &nd the derivative 
of the variable. 

25. Proof of I. This follows immediately from the definition of 

\w dx 

a derivative. For, since — = 1, its limit — = 1. 
Ax dx 

26. Proof of II. A constant is a quantity whose value does not 

vary. 

Hence Ac = and — = j therefore its limit — = 0. 
Ax • dx 



28 DIFFERENTIAL CALCULUS 

27. Proof of III. Let y = u-\-v, and suppose that when x re- 
ceives the increment Ax, u and v receive the increments Au and Av, 
respectively. Then the new value of y, 

y + Ay = u + Au + v + Av, 

therefore Ay = Au-{- Av. 

Divide by Ax ; then 

Ay _Au Av 
Ax Ax Ax ' 

Now suppose Ax to diminish and approach zero, and we have for 
the limits of these fractions, 

dy _du dv 
dx dx dx' 

If in this we substitute for y, u -{-v, we have 

d / . N du , dv 

— (u + v) = 1 . 

dx dx dx 

It is evident that the same proof would apply to any number of 
terms connected by plus or minus signs. We should then have 

d , , , , s du . dv , dw , 
dx dx dx dx 

28. Proof of IV. Let y = uv; 

then y + Ay = (u + Au) (y + Av), 

and Ay = (u + Au) (y + Av) — uv = vAu + (u -f- Au) Av. 

Divide by Ax ; 

then ^/ = „^ + ( M + Au) ^. 

Aaj Ax Ax 

Now suppose Ax to approach zero, and, noticing that the limit of 

u + Ait is u, we have 

dy du , dv 

-£ = v \-u — ; 

dx dx dx 

,i d / k du , dv 

that is, — (uv)=v \-u — . 

dx dx dx 



DIFFERENTIATION 29 

29. Product of Several Factors. Formula IV. may be extended to 
the product of three or more factors. Thus we have 

d , v d , x d , •. , dw 

— ( uviv) as — (uv -w)=w — (uv) + UV 

dx K ) dx y } dx K J dx 



du , dv\ . dw 

= w[ v \-u — ] -{-uv 

dx dx dx 



du , dv , dw 

= vw \-uw [-uv — . 

dx dx dx 



It appears from the preceding that the derivative of the product 
of two or three factors may be obtained by multiplying the deriva- 
tive of each factor by all the others and adding the results. 

This rule applies to the product of any number of factors. To 
prove this, w& assume 



d ( \ du, du„ . 
— I Wh ••• u n j = u 2 u 3 ••• u n — \ -f u x u z u A ... u n — - H h u x u 2 ~-u, 

Then ^-( l( \ u -2 — w B w n+1 J = u n+l -—( UjU 2 • •• uA + u&v ••• u n 



du n 
dx 

du„^ 



dx 

du* , duo . du n 

= u 2 u 3 • • • u n+1 — i + Uju 3 u 4 • • • u n+l —? H p. w^ • • . w n _!W n+1 — a 

ax ax dx 

+ ? , lW2 ... Wn _n±i. 
ax 

Thus it appears that if the rule applies to n factors, it holds also for 
7i-|-l factors, and is consequently applicable to any number of 
factors. 

Tlie derivative of the product of any number of factors is the sum of 
the products obtained by multijjlying the derivative oj each factor by all 
the other factors. 



30 DIFFERENTIAL CALCULUS 

dc 
30. Proof of V. This is a special case of IV., — being zero. But 

dx 

we may derive it independently thus : 

y = cu, 

y + Ay = c(u + Aw), 

Ay = cAw, 

Aw Aw 
— = c — ? 

Ax Ax 

dy du d, , x du 

-^ = c — , or — (cw) = c — 
dx dx dx dx 



31. Proof of VI. ~Lety = u 



Then y + Av 



w + Aw p 
v+Av ' 



therefore A y = «±* tt - ? = »* M ~ "f » 

Aw Av 
v w — 

and Aw_ ^ x ^ x 

Ax~ (y + Av)v 

Now suppose Ax to approach zero, and noticing that the limit of 
v + Av is v, we have 

dw dv 

V u 

dy _ dx dx 

dx ~~ v 2 
Or we may derive VI. from IV. thus : 



Since y = -, 

v 



therefore yv = u. 



DIFFERENTIATION 31 

t> ttt dy . dv du 

By IV., v-f + y— = — , 

dx dx dx 

dy __ du u dv, 

dx dx v dx 1 

du dv 
v u — 

therefore dy _ dx dx 

dx v 2 

32. Proof of VII. First, suppose n to be a positive integer. 

Let y = u n , 

then y + Ay = (u + Aw) n , 

and Ay = (u + A?*)' 1 — u n . 

Putting u' for u + A?<, we have 

Ay = u" — u n = (V — u) (u" l ~ l + u hl ~ 2 u + «4 ,B ~ 3 w 2 H + u n ~ l ), 

that is, Ay = Au (u'*- 1 + it" 1 - 2 u + u*^V f- w*- 1 ), 

^ = (u*- 1 +u' n - 2 u + m'- 8 ^ 2 ... + a- 1 ) — . 

Ax N ' Ax 

Now let Ax diminish ; then, u being the limit of u', each of the n 
terms within the parenthesis becomes u n ~ l ; therefore 

-/ = nu n ~ l — • 
ax dx 

Or it may be proved by regarding this as a special case of Art. 29, 
where u l7 u 2 , ••• and u n are each equal to u. 

Then — (O =u n ~ lC — +u n - 1 du + ... to n terms 
dx dx dx 






dx 



Second, suppose n to be a positive fraction,-?. 



Let y = u", 

then if = u p ; 

therefore — Of) = — (u p ). 

dx KJ J dx K } 



32 DIFFERENTIAL CALCULUS 

But we have already shown VII. to be true when the exponent is 
a positive integer ; hence we may apply it to each member of this 
equation. This gives 

qyq ijL =pvP i . 
dx dx 

therefore 4=«!^* 

dx q y q ~ l dx 



2 
Substituting for y, u q , gives 



dy _p u p ~ l du _p P q~ l du 

dx q p _pdx q dx' 
u q 



which shows VII. to be true in this case also. 



Third, suppose n to be negative and equal to — m. 

Let y = u~ m =: — ; 

u m 

i T7T dy dx dx m ,du 

by VI., /- = s- — = — = - mic m - 1 — 

dx u 2m u 2m dx 

Hence, VII. is true in this case also. 



EXAMPLES 

Differentiate the following functions : 
1. y = x\ 

■ * = !(*•). 
dx dx 

If we apply VII., substituting u = x and w = 4, we have 

— (z 4 )=4ar } — =4ar>. by I. 
dx K J dx J 

Hence, ^ = 4^. 

dx 



DIFFERENTIATION 33 



2. y = 3» 4 + 4aj 8 . 



*? = J*(3a;*+4a?)= — (3 x*) +-^(4 a 8 ), 
by III., making u = 3x 4 and u = 4# 3 . 

|(3^)=3|(A byV. 
= 3-4^ = 12^. 
Similarly, — (4 x 3 ) = 4 — (or) = 4 . 3 x 2 = 12 a?. 

Hence, ^= 12a 3 + 12« a = 12 (a? + a? 2 ). 

e2as 



3. y = ** + 2. 



dx dx dx 

|(,f) = |,i ; by VII. 

|(2)=0, by II. 



Hence, ^ = 1** 

dx 2 



4. y=3Vx-A + | +a . 



dx dx y J dx K J dx K ' dx 



3 -I 
= 2* ^ 


- 2 (- 


_1 


--*- + 

2x* 


_1 _ 


3 

x 4 ' 



34 DIFFERENTIAL CALCULUS 



5. y = %£ 

ar + 3 



dy_ d / x + 3 \ 
dx dx\x 2 -{-3 



Applying VI., making 

u = x + 3 and v = x 2 + 3, we have 

dfx+3\ (- 2 + 3)|(, + 3)-(, + 3)|(^ + 3) 



<toV<s 2 + 3y (x 2 + 3f 

_ f + 3-(x + 3)2x = 3-6a;-^ 
(z 2 + 3) 2 ~~ (> 2 + 3) 2 

Hence, dy = S-6x-a* m 

9 dx (x 2 + 3) 2 

6. 2/ = (x 6 + 2)t. 



dx dx 



If we apply VII., making 



2 
^ = x 2 + 2 and w = -,we have 

o 



Hence, 

dx 3(x 2 + 2)3 



3 3(» 2 + 2)i 



7. y = (x 2 + l)Vx*-x. 



^=A[(aj» + l)(aJ»-aOH 






DIFFERENTIATION 35 

If we apply IV., making 

m = as? + 1 and v = (x 3 — x)-, we have 

£[(rf + i)(#-.)»] 

= (a- 2 + 1) |- (X s - »)* + (*" - K)i|- 2 + 1). 



(x 2 + l)=2x. 



dx 2 dx 

dx 



Hence ^ = i(.T 2 + 1)(3 x 2 - 1) (ar 3 -*)-£ + (s 3 -»)*2a 

= (j 2 + 1) (3 x- 2 - 1) + 4 afo 8 - a;) = 7 a; 4 - 2 a; 2 - 1 
2 (.r 3 - xft 2{x z - a>)* 



8. ^ = 3x 10 -2a; 6 + ar ? -5, ^ = 3 (10 a 9 - 4 z 5 -far 2 )- 

10. y = (x + 2a)(x-a) 2 , ^=3 (a 2 -a 2 ). 

, , ,11.. <fy 4 (a; 3 — q3)3 

11. s,= (»*-ai)« S = 3ic f 

Differentiate Example 11 also after expanding. 

12 v= 2x ~ 1 d y = 2x 

' J (x-if dx (x-iy 

13. y = z(a*+5)*, g = 5 (ar + l)(x 3 + 5)i 



14. y = 



DIFFERENTIAL CALCULUS 

x dy a 2 



Va 2 - x 2 ' dx (a 2 - x 2 f 

15 (^-a 2 )^ dy = 3aWx*-a 2 



16 



la — x 



dx 



dy 



x dx 2 xy/ax — x 2 

Differentiate both members of the identical equations, Exs. 17-19. 

17. (x 2 + ax -h a 2 )(x 2 — ax + a 2 ) = x* -f aV + a 4 . 



18. /'^-±A 2 Y = ^ + 2a 2 + a 4 »- 2 . 



19. , 2 ^_ = 1+ 2 



20. a = * (* 2 + <*)*& f = (nt 2 + a 2 )(* 2 + a 2 )T 



21 ^^ ( 2 ^ 2 ~ 3 ) 3 ■<fy_6( 3* 2 +4Q(2; 2 -3) 2 

' y (? + 2) 2 ' dt (? + 2f 

* y (2a-3a) 9 ' dx (2a-3a?) 10 

23. 2,= (z + l) 3 (3z-8) 4 (z + 2^ 

^ = 3 (13 a 2 - 24) (a; + 1) 2 (3 a; - 8) 3 (x + 2) 5 . 

N n-i dy_ n(x n + l) 

24. <, = *(*» + n)-> ^- (ajn + n) l 

25. </= aJ_a d2/ - a ' 



V2 aa; - oj 2 *» (2 aa - x 2 )% 



DIFFERENTIATION 37 

*"• * ~~ ~4 ' ,7^ Q ^.4 



rf x 3 x 4 



cty 



28. *=(*^2)JJ7i; g= 3 t 
\ d* 4 (ti + 1)1 



29. t, 



_ (x 3 — «?)» c7y _ 2 a 3 x 2 





if l' 


d* (. r 3_ h a 3 )V-«') f 


30. 


y = ^jX- ~X-\-l 

V \^+x + l' 


cfy_ x-'-l 




ax (a* +x + 1) Vx 4 + x 2 + 1 


31. 


6 x 2 4- 6 * + 1 
(4*+l) f 


dy_ 12 or 9 


*■ (4« + l) f 


32. 


2/== ( iC 2_3ax)^(4^ 


+ 8az + 15« 2 )3 ) 

c7?/_ 4(2 x 3 - 9a 3 ) 

*b (rf _ 3 cu.) 5(4 rf + 8 ax + 15 a 2 )* 



33. y = (x + Vx 2 4- l) n (" Vx 2 + 1 - *), 

cZ.V 



te = (,r-l)(z+Vz 2 + l)». 

34. For -what values of x is 3 x 4 — 8 x 3 an increasing or a decreas- 
ing function of x ? 

Arts. Increasing, when x > 2 ; decreasing, when x < 2 

35. A vessel in the form of an inverted circular cone of semi- 
vertical angle 30°, is being filled with water at the uniform rate of 
one cubic foot per minute. At what rate is the surface of the water 
rising when the depth is 6 inches ? when 1 foot ? when 2 feet ? 

Arts. .76 in. ; .19 in. ; .05 in., per sec. 



38 DIFFERENTIAL CALCULUS 

36. The side of an equilateral triangle is increasing at the rate 
of 10 feet per minute, and the area at the rate of 10 square feet per 
second. How large is the triangle ? Ans. Side = 69.28 ft. 

37. A vessel is sailing due north 20 miles per hour. Another 
vessel, 40 miles north of the first, is sailing due east 15 miles per 
hour'. At what rate are they approaching each other after one 
hour ? After 2 hours ? Ans. Approaching 7 mi. per hr. ; separating 
15 mi. per hr. 

When will they cease to approach each other, and what is then 
their distance apart ? 

Ans. After 1 hr. 16 min. 48 sec. Distance = 24 mi. 

38. A train starts at noon from Boston, moving west, its motion 
being represented by s = 9 f. From Worcester, forty miles west of 
Boston, another train starts at the same time, moving in the same 
direction, its motion represented by s' = 2 f. The quantities s, s', 
are in miles, and t in hours. When will the trains be nearest to- 
gether, and what is then their distance apart ? 

Ans. 3 p.m., and 13 mi. 
When will the accelerations be equal ? 

Ans. 1 hr. 30 min., p.m. 

39. If a point moves so that s = y7, show that the acceleration 
is negative and proportional to the cube of the velocity. How is 
the sign of the acceleration interpreted ? 

40. Given s = - + bt 2 ; find the velocity and acceleration. 

41. A body starts from the origin, and moves so that in t seconds 
the coordinates of its position are 

o 

Find the rates of increase of x and y. 

Also find the velocity in its path, which is 
ds 



vdHty 



vn , , Ans. 5f + o. 



DIFFERENTIATION 39 

42. Two bodies move, one on the axis of aj, and the other on the 
axis of y, and in t minutes their distances from the origin are 

x 5= 2 f 2 - 6 t feet, and y = 6 1 — 9 feet. 

At what rate are they approaching each other or separating, after 
1 minute ? After 3 minutes ? 

Ans. Approaching 2 ft. per min. ; separating 6 ft. per min. 
\Yhen will they be nearest together ? Ans. After 1 min. 30 sec. 

43. In the triangle ABC, L and M are the middle points of BC 
and CA respectively. A man walks along the median AL at a uni- 
form rate. A lamp at B casts his shadow on the side AC. Show 
that the velocities of the shadow at A, M, C, are as 2 2 : 3 2 : 4 2 ; and 
that the accelerations at these points are as 2 3 : 3 3 : 4 3 . 

Suggestion. — P being any position of the man, draw from L a 
line parallel to BP. 

33. Formulae for Differentiation of Logarithmic and Exponential 
Functions. 

du 

VIII. -^-log u « = log a e^. 
dx u 

du 

TV d i dx 

IX. — log e w=— • 
dx u 



X. — a u = log e a-a u — • 
dx dx 



XI. *e" = e v — 
dx dx 



YTi d .du , -, v dv 

XII. - "='// ■ h log e ?^ • u v — • 

dx dx dx 



40 DIFFERENTIAL CALCULUS 

34. Proof of VIII. Lety = log a u, 
then y + Ay = \og a (u + Au), 

Ay = \og a (u + Aw) - log a u = lo go ^±^ 

= log/l + ^=^log„(l+^ f ". 

\_ u J u \ u J 

Dividing by Ax, 

^logA+^f^ CD 

Ax \ u J u 

If Ax approach zero, Au likewise approaches zero. 



Now Lim^^l + ^Y^Lim^Jl + l 



For, if we put — = z, 

Au 



( i+ fr=( i+ t: 

and as An approaches zero, z approaches infinity. 
But in Art. 12 we have found 
Lim^M +-j =e; 

therefore Lim AM=0 [ 1 -\ — - r" = e. 

Hence, if we take the limit of each member of (1), 

du 
dy i dx 
dx u 



DIFFERENTIATION 41 

35. Proof of IX. This is a special case of VIII., when a = e. 

In this case 

log a e = log e e = l. 

Note. — Logarithms to base e are called Napierian logarithms. 
Hereafter, when no base is specified, Napierian logarithms are to be 
understood; that is, 

log u denotes log e u. 

36. Proof of X. 

Let y = a u . 

Taking the logarithm of each member, we have 
log y = u log a; 

dy 

therefore by IX., dx , du 

J 7 — = loga — 

y dx 

Multiplying by y = a u , we have 

dy t u du 

— = loga • a u — . 
dx dx 

37. Proof of XI. This is a special case of X., where a — e. 

38. Proof of XII. Let y = u v . 

Taking the logarithm of each member, we have 
\ogy=v\ogu; 

dy du 

therefore by IX., dx dx . n do 

J ' — = + logw— . 

y u dx 

Multiplying by y = u v , we have 

dy r-idu . i v dv 

-?- — vu v l hlog^-^ v — 

dx dx dx 

The method of proving X. and XII. by taking the logarithm of 
each member, may be applied to IV., VI., and VII 
This exercise is left to the student. 



42 



DIFFERENTIAL CALCULUS 



EXAMPLES 

(See note, Art. 35.) 



1. y = log (2^ + 3 r% 



2. y = x n log (a# + 6), 



3- y 



xlogx 
4. 2/ = log 10 (3 x + 2), 
ax — b 



5. # = log 



aa? + 6 



I. y = lo{ 



3^ + 1 

' «+3 



8. y = a x e x , 

9. y = log(a* + 6*), 
10. y = (««-_ 1)*, 



dy _ 6 (a; + 1) 
dx 2 a? 2 + 3 x 



dy 
dx 



ox 



ax+b 



+nlog(ax+b) 



dy _ _ 1 + log x 
dx (x log x) 2 

riy = 3 log 10 e = 1.3029 

da 3x + 2 3a+2 

dy _ 2 ab 



dx a 2 x' 2 — b 2 



dy 



dx 3a 2 + 10a;-f 3 

dy _ 2 (f- - 1) 
eft ^4-^-j_^ 

|2=(l + loga)a*e". 

dy _ a x log a + &* log b 
dx ~ a x + b x 



dy 
dx 



= 8 e 2x (e 2x - l) s . 



Differentiate Ex. 10 also after expanding. 



11. y = 6* + V », 

x — 3 



<fy = 24a--10^ & 
cte (x-S) 2 



12. y=(3x-l) 2 e 3x - 2 , 



dy 
dx 



3 (9 x 2 - l)e 3x -\ 



DIFFERENTIATION 43 



13. y = x 5 5 x , 




~JL = x 4 o T (o + x log 5). 
da; 


14. y = log log x - 


1 

log x 


d# _ 1 + log X 

dx x (log a;) 2 



Differentiate both members of the identical equations, Exs. 15-18. 

15. (x + e x y = .r 4 + 4 .rV + 6 xV" + 4 a* 3 * + e 4 *. 

16. (a* - e x f = a 3 " - 3 cr*e x + 3 a x e 2x - e*. 

17. log (e 2 * + e' 2a ) = log (e x ~ a + e a " x ) + x + a. 

18. .i Jog a = a log x . 

.. Q x los: x i / , . ^ v dy log x 

20. y = log (Vx~HTa + Vx), 



21. y = log (2a-+V4x*-l), 



<ty_ 


1 




dx 


2Vx* + 


ax 


d y _ 


2 




da 


V4r>- 


1 


dy _ 


1 





22. y = log ^±i-^ 

a.,- + 1 + 1 ax xVx + 1 

23. y = x [(log x) 2 - 2 log x + 2], 4 = (log X) 2 - 

UX 

oa i / "'•' / - (x + a)* + (x - a)* 

24. r = log(VJ+7-V—,fc S — . ^ f ..'■ 



25. ,, = log(V* + 3 + V* + 2)+V(* + 3)(z + 2), g=^/|±| 

26 ■i = W- ,r - 1 + *'"' , •'■'! = 2 „(,■■■'-?-<) 

'.-"' + 1 + <r" dx c-"' + 1 + e"-" 



44 DIFFERENTIAL CALCULUS 

87. y = log lo « x , ®L = 1 . 

1 + log x dx x log a; (1 + log x) 

, dv 30 e°* 



2/ = 3 log ( Vx 2 + 3 -3) + log ( Vx 2 + 3 + 1), 



30. y = log (a + V2ax-a 2 ) + 



fly __ 4 a; 

^"~^-2V^T3 



Va 4- V2 a; — a 

0> 1 



dx 



2(x + V2 ax - a 2 ) 



31 v _,i og a;2 + l4-V^ + 3a ; 2 + l j dy = x 2 -! 

x dx x^/x^ + Stf + l 

The following may be derived by XII. or by differentiating after 
taking the logarithm of each member of the given equation. 

33. y = x nx , ^| = nx™ (1 + log x). 

34. y = (ax>) x , ^= (ax 2 ) 1 [2 + log (ax 2 )]. 

ax 

35. 2/ = x** 2 , ^ = ax°* 2+1 (l + 21ogx). 

ax 

36. 2/ = (logx)*, ^ = (log x)/-i— + log log x V 

ax ° Vlogx y 

37. y = x< 10 * ^ * te (w + 1} (log ,,) V io, <^ 



38 . ,»(-£_£ * rf f-fL,W-i-+il€g^-A. 

\x + aJ dx \x-\-aJ \x + a a x + a) 



DIFFERENTIATION 45 

The method of differentiating after taking the logarithm of the 
expression may often be applied with advantage to algebraic func- 
tions. This is sometimes called logarithmic differentiation. 

In this way differentiate Exs. 21-26, pp. 36, 37. 

I X 

39. Find the slope of the catenary y = ~(e a -f- e a ), at x = 0. 

What is the abscissa of the point where the curve is inclined 45° 
to the axis of X ? Ans. x = a log e (1 + V2). 

40. When does log 10 # increase at the same rate as #? 

Ans. When x = log 10 e = .4343. 
When at one third the rate? Ans. When x = 1.3029. 

Verify these results from logarithm tables. 

41. If the space described by a point is given by s = ae c + be~ f , 
show that the acceleration is equal to the space passed over. 

42. If a point moves so that in t seconds s = 10 log feet, 

£ + 4 
find the velocity and acceleration at the end of 1 second. At the end 
of 16 seconds. Ans. Velocity = — 2 ft., and — .5 ft. per sec. 
Acceleration = .4, and .025. 

43. For what values of a? is y = log (x - 2) 3 - 9 ^~ 36 x + 32 
an increasing or a decreasing function ? \ ~ ) 

Ans. Increasing when x > 3; decreasing when x < 3. 

39. Formulae for Differentiation of Trigonometric Functions. In 

the following formulae the angle u is supposed to be expressed in 
circular measure. 

d du 

XIII — sin w = cos u—- 
dx dx 

XIV —cos u = - sin u— . 
dx dx 

XV. — tanw = sec 2 M— . 
dx dx 



46 DIFFERENTIAL CALCULUS 



d , o du 

XVI — cot u = — cosec 2 w — • 
dx dx 

XVII. — see u= sec u tan % — 
dx dx 

XVIII. — cosec u = — cosec u cot ?t — - 
dx dx 

XIX. — -versw=sm?t— • 
ax ax 



40. Proof of XIII. Let y = sin u, 

then y-\-Ay = sin (it + Am); 

therefore A y = sin (w -f- A u) — sin ?t. 

But from Trigonometry, 

sin A - sin B = 2 sin* (A - B) cos \ (A + B). 



If we substitute ^4 = u + A w, and B = m, 

we have Av = 2cos(mH — -)sin— - 



9 

Aw 



Hence, H = cos u + *g) l^ 

Ax V 2 An Ax 



Now when Ax approaches zero, Aw likewise approaches zero, and 
as Au is in circular measure, 

• Au 
sin — 
2 
Lim AM=0 -^- = 1. See Art. 12. 

T 

TT dy du 

Hence, — = cos a 



dx dx 



DIFFERENTIATION 47 

41. Proof of XIV. This may be derived by substituting in XIII. 



for u, £-?/. 



Then |-(l-«) = cos(|-„)|(| r «) 

d ■ ■ f du\ d» 

or — cos?* = sin ?/[ =-sm« — 

dx V dxj dx 



42. Proof of XV. Since tan u = ^-^, 

cos u 

tf . d 

cos ?/ — sm v — sm ?/ — cos u 
. „ T tf ete (to 

bv \ 1., ~y~ tan " = 5 

d* cos- >/ 

, dn . . o dv du 
cos 2 b— t sm- u — — 



sec- " 

CfeB 



COS" ?' COS" w. 



43. Proof of XVI. This 'may be derived from XV. by substitut- 



ing — — v for ?/. 



44. Proof of XVII. Since sec u = — 



cos u 



d . c?u 

cos u sm w — 

, „, f? eta dx 

bv A L, 3~ sec u = 5 = = 

J 7 ax cos- ?< cos- u 

. du 

= sec v tan >/ — 
dx 

45. Proof of XVIII. This may be derived from XVII. by sub- 
stituting ^ — u for u. 

46. Proof of XIX. This is readily obtained from XIV. by the 

relation 1 

vers u= 1 — cos u. 



48 DIFFERENTIAL CALCULUS 



EXAMPLES 

1. y = 3 sin 3x cos 2x — 2 cos Sx sin 2x, -^ = 5 cos 3x cos 2x. 
u dx 



2. y = log cos 2 x-\-2x tan a? — a? 2 , -^-= 2x tan 2 #. 

3. ?/ = log (sec ra# -f tan mx), — = m sec wa?. 

C*3J 

>i i / • 2 i * 2 \ dy 2 (a—b) tan # 

4. ?/ = log (a sm 2 cc + 6 cos 2 a), — = — r 1 

u 5 v ; dx a tan 2 x + b 

c i //) \ , a • dv sin 

5. y = cos a log sec (8 — a) + sin ot, -^ = — . 

dv cos (0 — a) 

6. y=(m — 1) sec m+1 #— (ra + l)sec m_1 a;,^==(m 2 — l)sec m-1 a;tan 3 a;. 

cia; 

7. 2/ = log tan ( aa? — ^"j, -^ =— 2a sec 2ax. 

8. r = log [sec tan 6 (sec + tan 0) 2 ] , — = (sec 6 + tan ^ 2 - 

& L v y J ' dB tan 

9. 2/ = cosec m ax cosec" bx, 

— ^ = — cosec™ ax cosec" bx (ma cot ax + n& cot bx). 
dx 

10. w = 2x 2 sin 2x + 2x cos 2# — sin 2x, -&■ == 4a 2 cos 2a\ 

da; 

11. ?/ = 2 tan 3 a; sec a; + tan a; sec x — log (sec x + tan #), 

-^ = 8 tan 2 x sec 3 a;. 
dx 

- ' sin a; -f cos x dy 2 sin x 

12 - »- ? ' Tx V 

13. y = e 3 '(sin 2a; - 5 cos 2a;), ^ =13e to (sin2x - cos2z). 

ax 



14. y = log 



DIFFERENTIATION 49 

cos x dy sin a 



cos (x + a) dx cos x cos (x ■+■ a) 



15. y = sin 3 4.r cos 4 3x, -^ = 12 sin 2 4a; cos 3 3a; cos 7x. 

dx 

, « t sin x 4- vers x dy 

16. y = log ! , -^ = secx. 

sin x — vers a; dx 

17. 2/ = (sin 2 a;) 1 , — = 2/ (log sin 2x + 2a; cot 2a;). 

da; 

18. ?/ = (tan a;) 8inz , — = y (cos a; log tan a; + sec a?). 

dx 

19. y = (sin x) log c0 * x , -^ = y (cot x log cos a; — tan x log sin x). 

dx 

20. 7/ = tana;seca; + log J^L±4^, - = 2 sec 3 x. 

* 1 — sin a; da; 



21. 7/ = (tan a; -3 cot a;) Vtana;, dy = 3sec 4 a; 

d * 2 tan* x 

sin^fl — «) 



22. y = log 



dy _ sine* 
sini^ + a)' d0~"cos«-cos0 



23. y = a log (a sin a; + 6 cos a;) + 6a, ^ = _^_±&!_, 

da; a tan a; -f b 



24. y = - 



sin 



( 2 - + i) 



tan- -2 



25. y = log 



da; 1-|- sin 4a; 



dy 3 



2tan?-l d * 4-5 sW 



50 DIFFERENTIAL CALCULUS 

oa a sin if + b vers x dy 2 ab vers x 

«D. ]) = ; , — - — ; — 

a sm x — b vers x dx (a sm x — b vers x)~. 

In each of the following pairs of equations derive by differentia- 
tion each of the two equations from the other: 

27 . sin 2 a; = 2 sin x cos x, 
cos 2 x = cos 2 a? — sin 2 ic. 

2 tan x 



28. sin 2 a; 

cos 2 a* 



1 -f tan 2 aj' 
1 — tan 2 aj 



tan 2 # 



29. sin 3 a; = 3 sin a? — 4 sin 3 a:, 
cos 3 x — 4 cos 3 a; — 3 cos #. 

30. sin 4 a? = 4 sin a; cos 3 a; — 4 cos a; sin 3 #, 
cos 4 a: = 1 — 8 sin 2 x cos 2 x. 

31. sin (m + n) x = sin mx cos rase + cos mx sin wa;, 
cos (m -\-n)x= cos mx cos wa; — sin mx sin wa\ 

32. If 6 vary uniformly, so that one revolution is made in ir sec- 
onds, show that the rates of increase of sin 0, when = 0°, 30°, 
45°, 60°, 90°, are respectively 2, V3, ^/% 1. 0, per second. 

33. If 6 is increasing uniformly, show that the rates of increase 
of tan 0, when = 0°, 30°, 45°, 60°, 90°, are in harmonical progres- 
sion. 

34. For what values of 0, less than 90°, is sin 6 -f cos an increas- 
ing or a decreasing function ? 

Find its rate of change when = 15°. Ans. 

vr 

35. The crank and connecting rod of a steam engine are 3 and 10 
feet respectively, and the crank revolves uniformly, making two 
revolutions per second. At what rate is the piston moving, when 



DIFFERENTIATION 51 

the crank makes with the line of motion of the piston 0°,. 45°, 90°, 

135°, 180° ■: 

If a, b, x, are the three sides of the triangle, and $ the angle 
opposite b, 

x = a cos + V6 2 — a- sin 2 6. 

Ans. 0, 32.38, 37.70, 20.90, 0, ft. per sec. 

36. A crank OP revolves about O with angular velocity <o, and a 
connecting rod PQ is hinged to it at P, whilst Q is constrained to 
move in a fixed groove OX Prove that the velocity of Q is w. OP, 
where R is the point in which the line QP (produced if necessary) 
meets a perpendicular to OX drawn through O. 

47. Inverse Trigonometric Functions. The inverse trigonometric 
functions are many-valued functions; that is, for any given value of 
x. there are an infinite number of values of sin -1 #, tan -1 #, &c. 

For example, sin -1 -= ±-± 2mr, where n is any integer. 

But if the angle is restricted to values not greater numerically 
than a right angle, sin -1 x will have only one value for a given value 

of x. Then sin -1 - = -, sin _1 ( ] = — - • We thus regard sin -1 x, 

2 6' V 2y 6 

cosec -1 x, tan -1 x, and cot _1 .T, as taken between — - and -, that is, in 

the first or fourth quadrants. 

But cos _1 .r, sec -1 x, and vers -1 a;, must be taken between and v, 
that is, in the first and second quadrants, which include all values of 
the cosine, secant, and versine. 

These restrictions are assumed in the following formulae of differ- 
entiation. 

48. Formulae for Differentiation of Inverse Trigonometric Functions. 

flu 
XX. — sin 



dx VI - u 2 

clu 
XXI. i^cos-S< = 



dx x i 



52 



DIFFERENTIAL CALCULUS 



XXII. 



XXIII. 



XXIV. 



XXV. 



XXVI. 



du 
dx 



dx 


~l + w 2 




d , i 
— cot u 
dx 


du 
dx 

l+u 2 




— sec -1 u 


du 

dx 




dx 


«V« 2 -1 




d 

— cosec - 


du 
i ## dx 




dx 


u-Vu 2 - 


1 


d 

— vers 1 


du 

dx 
jl = • 





dx 



V2w 



49. Proof of XX. 

therefore 

By XIIL, ' 

therefore 
But 



Let y— sin -1 %; 
sin y = u. 

du du 
cos 2// = — ; 
dx dx 

du 

dy _ dx _ 
dx ~ cos y 



cos y = ± Vl- sin 2 2/ = ± VT 



If the angle y is restricted to the first and fourth quadrants 
(Art. 47), cos y is positive. 

Hence 



and 



cosy = 


Vl- 


■< 


dy _ 


du 
dx 





dx VI 



DIFFERENTIATION 53 

50. Proof of XXI. Let y = cos _1 w; 

therefore cos y — u. 

A , TT ^ ■ dy du 

dx dx 



therefore 



du 
dy_ dx 
dx~ smy 



But sin y = Vl — cos' 2 y = Vl — uK 

If the angle ?/ is restricted to the first and second quadrants 
(Art. 47), sin y is positive. 



Hence 


sin y - 


= VI - u 2 , 


and 


dy_ 
dx~ 


du 
dx 




■VI -u 2 


51. Proof of XXII. 


Let 


y = tan -1 u ; 


therefore 




tan y = u. 


By XV, 




o dy du 
sec^-^ = — ; 
dx dx 

du 


therefore 




dy _ dx 
dx sec 2 ?/ 


But 
therefore 




sec 2 y = 1 -h tai 

du 
dy _ dx 



dx 1 + u 2 

52. Proof of XXIII. This may be derived like XXIL, or from 
cot -1 u = tan -1 - . 



54 DIFFERENTIAL CALCULUS 

53. Proof of XXIV. This may be obtained from XXI. Since 

sec -1 u = cos -1 -, 
u 

d /1\ 1 du du 



d _! d _,1 dx \uj u 2 dx dx 

sec * u = — cos l - = 



dx dx u I i ./ i «V % 2 -1 

54. Proof of XXV. This mav be obtained from XX. Since 



1 • ll 

cosec -1 w = sin - -> 










u 












d 


f 1 ) 


1 du 


du 


d _! d 


, 1 dx 

n -1 — : — 


w 


u 2 dx 


dx 



^-i +-i 



,-\/ir — 1 



55. Proof of XXVI. This may be obtained from XXL Since 
vers _1 w = cos -1 (1 — w), 

d /-. _ v du 

vers -1 m = — cos x (1 — u) 



dx dx Vl-(l-w) 2 V2 



EXAM PLES 



! 4.-1 5® — 1 

1. y = tan x — - — > 

^ 9 



• 2/ = sec x — , 



3. y = sin x — - — > 
u - 7 



4. ?/ = vers -1 (8# 2 — 8# 4 ), 



5. 3/ = tan x > 



dt/ 


o 




da; 


5.r- — 2a? -h 1 




dy _ 


3 




dx 


^V4x 2 -9 




^.V _ 


1 




dx 


V(aj - 5)(2 - 


X) 


dy _ 


4 




da; 


VI - x 2 




% _ 


a 





DIFFERENTIATION 55 

6. y = tan * (3 tail 0), 

7. ?/ = sec -1 sec 2 0, 



9 

8. y = vers -1 



r + 1 

9 . y=cot -i *"+':' 



er* — e 



10. // = cosec -1 



dy _ 


3 


d6 


o - 4 cos 2 


dy_ 


2 


dO 


Vsec 2 + 1 


dy _ 


2 


dx 


*» + l' 


dy 


2a 


dx 


e 2ax_|_ e -2«x 


dy _ 


1 



* - 1 *c a V8ar + 2b - 1 

11. // = tan x (- cot ' > -^ = 0. 

5 6x + 1 da; 

12. y = cos -1 Vvers sc, — = — 4 Vl 4- sec a7. 

13. y = o tan- ? - b tan- * * = (a'-ftV , - 

J a 6 da; (a; 2 4-a 2 )<> 2 + & 2 ) 

14. >/ = cot- ^ + ^ , 

fa; — a 



15. 


• _i Sill X — COS SB 

y = sin A 

V2 


16. 


y = sin ■ ; , 



17. j/ = tan -1 (sec x 4- tan a?) , 



18. y = sin -1 - 



dy _ 




a 






da; 


a; 2 


+ a 




dx 


1. 


1 






f ty _ 


k 2 


-6 2 


da; 


&a 


• + ( 


-a; 2 ' 


•#_ 


.1 








da* 


2 









dy 



e' 4- e - * dx e x 4- e~ x 



19. y = cot ' 1 .,- - x 4- 1) - cut" 1 (a; - 1) 



dy 



da; a; 2 4- 1 



56 DIFFERENTIAL CALCULUS 

20. y=:tan- l4 + 5tanfl; ? * 



3 da? 5 + 4 sin 2a; 



21. 2/ = cos- 1 ^^-2 x /^ % = V6av-a^ 

22. 2/ = ^ 2 sec- 1 --2V^ = l, ^/ = 2 a; sec" 1 -• 

2 da; 2 

Differentiate both members of the identical equations, Exs. 23-28. 



23. 2cos- 1 ^±i = cos- 1 a. 

24. 3 vers- 1 x = vers" 1 [x(2x- 3) 2 ]. 



25. sin -1 x + sin -1 a = sin -1 (a y 1 — x z -\-x Vl — a 2 ). 

26. tan -1 mse +. tan -1 nse = tan -1 ^ — ^ • 

1 — mnxr 



ol - ,2 a; + 2 , _i /a; + 1 

27. vers * — =[ -—=2tan \ — ^ — 

» + 3 \ 2 

28. tan- "'tan*-* = ^ fa ^ \ _ ^ b . 

a6(l + tanx) \b J a- 

OQ 01 aj 2 -2a; + 5 14 . _!^-5 dy 12x 2 -20 

29 ' ^ = 2lQg ^ + 2, + 5 +tan 17^ & = 54^F+3' 

__ _ 4 a; — x s dy 4 

3 



Q1 . .^(a^-a 2 )^ dy 2\ / x 2 -a 2 

31. 2/=sm x — ^ — - — i— y -£ = — 

3 V3 a 2 a ^ a? V4 a 2 - or 

32. What value must be assigned to a so that the curve 

y = log e (x — 7 a) -f- tan -1 a«, 
may be parallel to the axis of X at the point x = 1 ? 



^4ns. J or — i. 



DIFFERENTIATION 57 

33. A man walks across the diameter, 200 feet, of a circular 
courtyard at a uniform rate of 5 feet per second. A lamp at one 
extremity of a diameter perpendicular to the first casts his shadow 
upon the circular wall. Required the velocity of the shadow along 
the wall, when he is at the centre ; when 20 feet from centre ; when 
50 feet ; when 75 feet ; when at circumference. 

Ans. 10, 9 T 8 g, 8, 6-f, 5 ft. per sec. 

56. Relations between Certain Derivatives. It is necessary to notice 
the relations between certain derivatives obtained by differentiating 
with respect to different quantities. 

To express — in terms of ' — • If y is a given function of x, then x 
dx dy 

may be regarded as a function of y. From the former relation, we 

have -^, and from the latter, — These derivatives are connected 
dx dy 

by a simple relation. 

It is evident that —£ = -—, 

Ax A# 

*9 

however small the values of Ax and Ay. As these quantities ap- 
proach zero, we have for the limits of the members of this equation, 

ft=i (i) 

dx dx w 

dy 

That is, the relation between -2 and — is the same as if they were 

-,. - .. dx dy 

ordinary tractions. a 

For example, suppose 

x=-^~. (2) 

y + 1 • w 

Differentiating with respect to y, we have 
dx a 

B 



dy G/ + 1) 2 

y( i), !=_M!=-!, by(2) . 



58 DIFFERENTIAL CALCULUS 

This is the same result as that obtained by solving (2) with refer- 
ence to y, giving 

y = - 1, 

x 

and differentiating this with respect to x. 

To express -^ in terms of -^ and — : that is, to find the derivative 
dx dz dx 

of a function of a function. If y is a given function of z, and z a 

given function of x, it follows that y is a function of x. This relation 

may often be obtained by eliminating z between the two given 

equations, but -^ can be found without such elimination. 
dx 

By differentiating the two given equations, we find —and—, and 

dz dx 

from these derivatives,— may be obtained by the relation 

dx J J 

dy_dy^dz^ /on 

dx dz dx 

For it is evident that — ^=— ^ — 

Ax Az Ax 

however small Ax, Ay, and Az. By taking the limits of the members 
of this equation we obtain (3). That is, the relation is the same as 
if the derivatives were ordinary fractions. 
For example, suppose 

y = * 5 , \ , 4 x 

z^tf-x 2 .} K) 

Differentiating these equations, the first with respect to z, and 
the second with respect to x, we have 

dz dx 

By (3), %L = bz\- 2x) = - 10x(a 2 - x 2 )\ by (4). 

dx 



DIFFERENTIATION 59 

The same result might have been obtained by eliminating z between 
(4), giving 

and differentiating this with respect to x. 

The relation (1) may be obtained as a special case of (3) by 
substituting y = x. This gives 






Another form of (3) is 



dx dz _ dx _ .. 

dz dx dx 


dy 


dx dy 

dz~dz' 


dx 


EXAMPLES 



(5) 



which is of frequent use. 



In Exs. 1-4, find — and thence — by (1). 
dy dx 

i ,- — a y ~ ^ dy _ (by — k) 2 __ bh — ak 

by — k' dx bh — ak (bx — a) 2 



2. x = vl +siny, 



3. x = 



1 + logy' 



dy _ 


_ 2 Vl + sin y 
cos y 


2 


dx 


V2-X 2 


dy. 


_(l-flog; V ) 2 _ 


f 


dx 


logy 


xy — x 2 

2z 2x 



4. a? = olog Vy + q " hV ^, dy = 2^/tf + cu / = e«-e 

Va dx a 2 

In Exs. 5-8 find ~ and — , and thence -j- by (3). 
dz dx dx J v ' 

9 2z-l' :;./;- 2' dx (a>+2) 2 ' 



60 DIFFERENTIAL CALCULUS 

6. y = log£±±, , = «■, dy = e-e~*_ 

z dx e x + e~ x 

7. y=ze z + e 2z , z = \og(x-x 2 ), ^/ = 4 a) 3 - 6 ar> + l. 

dec 

8. ?/ = log ! — , z = sec x -f tan #, 

62 -fa 

% a 2 — 5 2 



da? 2 ab -\- (a 2 -f 6 2 ) cos x 
9. Differentiate (a? 2 + 2) 2 with respect to sb 3 . 

n fi nrl 

dz 



(7 7/ 

Let y = (x 2 + 2) 2 , and 2 == ar 3 . It is required to find -j*- 



^ = 4aj(aj 2 + 2), — = 3a; 2 . 

cte dx 

B (5) dy = ±x(x 2 + 2) = ±(x 2 + 2) 

J w dz 3x 2 3x ' 



10. Find the derivative of — ; + — ; with respect to - + -■ 

a 6 x 6 c a x 



Ans. Sfc + l + £\ 



a; 2 / 

11. Find the derivative of sin 3a; with respect4o sin x. 

Ans. 3 (4 cos 2 x — 3). 

12. Find the derivative of taii -1 ^/^ with respect to log (1 + x). 

Ans. 



2-Vx 



13. Find the derivative of log — : with respect to 

a sin x — cos x 

j ab (a 2 tan x — b 2 cot x) 



a 2 sin 2 x — b 2 cos 2 x a 2 +b 

14. Given x = 5 cos <f> — cos 5cf>, y — 5 sin<£ — sin 5<£; find — . 

ciaj 

u.-i?t.s. — = tan 3 4>. 
dx 



CHAPTER IV 

SUCCESSIVE DIFFERENTIATION 

57. Definition. As we have seen, the derivative is the result of 
differentiating a given function of x. This derivative being generally 
also a function of x, may be again differentiated, and we thus obtain 
what is called the second derivative; the result of three successive 
differentiations is the third derivative; and so on. 

For example, if y = %\ 

dx 

-1^=12^, 

dxdx 

dx dx dx 



58. Notation. 

denoted by — ^« 
J dx 2 


The second derivative of 


That is, 


dry _ d dy 
dx 2 dx dx 


Similarly, 


dhj __ d d dy _ d d 2 y 
dx 3 dx dx dx dx dx 2 




d 4 y _ d d d dy _ d d 3 y 
dx 4 dx dx dx dx dx dx 3 




dry _ d d" _1 .y. 
dx n dx dx n ~ l 



01 



62 DIFFERENTIAL CALCULUS 

Thus, if 



y = 


■A 


dy_ 

dx 


= 4a 3 , 


dx 2 


= 12a;', 


fry = 
dx s 


= 24x. 



The successive derivatives are sometimes called the first, second, 
third, . . . differential coefficients. 

If the original function of x is denoted by f(x), its successive de- 
rivatives are often denoted by 

/'(•). /?(«)» /'"(•). /*(?). ••• /-(*)■ 

59. The rtth Derivative. It is possible to express the nth deriva- 
tive of some functions. 
For example, 

(a) From y = e ax , we have 

dy — ae ax fry _ a 2 e ax . . . fry = a"e ax . 
dec ' dx 2 dx n 



(b) From y = = (ax + b)~\ we have 

ax + b 

g = (-l)a(aa + 6)-*, g = (-1)( - 2)a\ax + &)-», 



g = (- 1)(- 2)(- 3)a 8 (aa + 6)~ 4 = (- 1)^3 a 3 (ax + 6)" 4 , 



2JL = (_ l)"|na-(aa! + 6)" Bj1 - — - 



dx 11 ' ■— y (ax + 6) 



n+l 



SUCCESSIVE DIFFERENTIATION 63 



(c) From y = sin ax, we have 

f -^- = a cos ax = a sin ( GKE-f- *" 
da; 



*fc = a 2 cos f aa + 1) = a 2 sin few + ^\ 



dx 






& = crcos f«e + — \= a 3 sin few +^\ 

(/"?/ - . / , ?nr\ 

—i = a n sin f ckb H ]. 

efo" 



EXAMPLES 

1. y = Za*- 5 rt + 20 ar» -or' + 2^ ^ = 120(0? - a? + 1). 

ax 

2 !/ = {x 2 -^, a ^M Q = 20(x 2 -l)(x 2 -4:)i 

ctx~ 

3. y = xr + x~ m , 

-■{ — m(m - l)(m — 2)aJ m " 3 — ra(m + l)(m + 2)ar ro - 3 . 

J ' da* L [3 |6 J9 

5. // = x 4 log a-*, 

6. 7 = a* 2 log (a* — 1), 



7. v = 4^_2)e*+(a,'-l)e 2x . 



8. ., = ^_3^ + — -3)e a , 



9. /• = log sec 0, 



da 5 


_24 

X 




cfy_ 


2(x 2 - 


-3a; + 3) 


dx"' 


(« 


-I) 3 


d 2 y_ 
dx 2 


= 4 x(e x 


+ *■)• 


d 2 x 
df 


:4t s e 2t - 




dfr 


- 6 sec 4 


0-4 sec 2 (9- 



64 DIFFERENTIAL CALCULUS 

10. y=ze- x (llsin2x-\-2eos2x), ^ = 125 e— sin 2 x. 

dx 3 

n . _, A 24a<l-a; 2 ) 

11. ?/ = tan 1 a?, — 2 = ^ >-> 

' dx A (1 + a; 2 ) 4 

12. ^tan-f^, g, 2(^ + ^-6). 
J 2 ' da; 3 ( e * + e -*)3 

,_ ,— a; d 3 y_2(x — aYx 2 + ■! ax + a 2 ) 

13. 2/ = logV^ + a- + t a n-'-, ^ (^ + a y 

14. y = (e a » + e-"*) sin a6, ^ + 4a 4 t/ = 0. . 

du 

15. ?/ = a;e x (sma; — cos a;)-}- 3 e x cos a?, — -^ = 4 a;e* cos x. 

16. 2/ = e- tan % ^ 2 + (tana;-l) 2 ^ = 0. 
l7 sinwg + coswa? ^ + ?^2/ + n 2 o. 

18. w = arYsin log a; + cos log x), x 2 — \ — 3 x-^- + 5 v = 0. 

dar da; 

19. 2/ = a 6 *, ^ = & w (loga) n a 6a! . 

(XX 

0* (—1Y- I 3 n \n — 1 

20. 2/ = log(3a;+2), ^ = ^ i- t==. 

9 , y da; n (3a; + 2) M 

, d"2/ (-l) w " 1 1.3.5...(2n-3) 

22. 2/ = sin5a;sin2a;, ^=y~3 w cos/3a;+ ^V ? n cosf7a;+^Yj 



SUCCESSIVE DIFFERENTIATION 65 

The following fractions should be separated into partial fractions 
before differentiating. 

1 fin.. ( — l) n \n 

23 i/ = - 1 



x 2 - 1 daT 



2 L(aj-1)" +1 (x + 1)"* 

oa 3a?-4 d"y , 1 . n| |~ 2 2" 

24. w = , — £ = (— l) n H — 

3 2.r-' + 3.r-2 da? K J L -\_( x + 2y +1 (2a>-l)» +1 

n 5 _ 13 ^1 = (- ±\n\ f 2 " +1 3" +1 



26. 



2x-- + x + l 
* 2.r'-x-l 



2 X + z 



-1 

On+1 



da" V ) L L3(x-l) n+1 3(2 a? + 1) 
x 2 dry _ (- l)" +1 4[w (x - n + 1) 



27. y= 

28. ^/ = 



(x + 2) 2 ' dx n (x + 2)' l + 2 

/ax + 1 \ 2 d" y 4 ( - i)" 01 " |* ( a:c + n ) 



\ax - 1/ dx" (ax — iy 



60. Leibnitz's Theorem. This is a formula for the ?ith derivative 
jf the product of two factors in terms of the successive derivatives 
3f those factors. 

A special case of Leibnitz's Theorem, when n = 1, is Formula IV., 

d , s du dv m\ 

T x {ttV) =Hx v+u <r x (1) 

For convenience let us use the following abridged notation : 

dv d?v d n v 

v i = — ? v 2 = — o> '" v n = 

dx dxr dx n 

du cPu d n u 

Uj = — , u 2 = — , ••• u n = — • 
dx " dx 2 dx" 



66 DIFFERENTIAL CALCULUS 



Then (1) becomes 
d 



(uv)=u 1 v+uv 1 (2) 



Differentiating (2), 

d 2 

— - (uv) = u 2 v + ihVi -h VqVi + uv 2 = u 2 v + 2 u l v 1 4- uv 2 , 
dx~ 

d 3 

— (uv) = u s v + U 2 V 1 + 2 iu>v x + 2u x V 2 + 0^2 4- uv 3 
dx> 

= u s v 4- 3 w 2 ^i 4- Sufis 4~ wy 3 . 

We shall find that this law of the terms applies, however far we 
continue the differentiation, the coefficients being those of the Bino- 
mial Theorem ; so that 

— (uv) = U n V + Wn-lVl + -^T^ Z Un-2» 2 -\ 1- UU^O^ + UV n . (3) 

(XX £ 

This may be proved by induction, by showing that, if true for 

d n d n+l 

— (uv), it is also true for (uv). This exercise is left for the 

dx nK J dx n+iy J 

student. 

In the ordinary notation (3) becomes 

d n f N d n u , d n ~ l udv . n(n-l)d n - 2 ud*v , 

(uv)= v + n h — h ••• 

dx nK } dx n dx n ~ l dx [2 dx n -*da? 

. dud n ~ l v , d n v 
dx dx"* 1 dx n 



EXAMPLES 

1. Given y — X s sin 2 x ; find by Leibnitz's Theorem — ^. 
9 dx 4 

d* 
From (3), — - (uv) == u#) + 4 u^ + 6 u 2 v 2 + 4 w^ 4 ttv 4 . 

Cta? 

^ = a; 3 , i<! = 3 a? 2 , u 2 = 6 a?, Wg = 6, u 4 = 0. 
v = sin 2 a?, %\ — 2 cos 2 x, v 2 = — 4 sin 2 x, v 3 — — 8 cos 2 a? 5 
v 4 = 16 sin 2 a;. 



SUCCESSIVE DIFFERENTIATION 67 

!&/ = <P , .r sin 2 x) = 0. sin 2 x + 4-6- 2 cos 2 .r -f 0-6 a; (- 4 sin 2 ») 
■^ dx 4 

-J- 4 • 3 x 3 (— 8 cos 2 «) + a? 3 16 sin 2 aj 

- 16 [(.i- 3 - 9 x) sin 2 x + (3 - 6 a 2 ) cos 2 »]. 

2. Given y = *«■*• find 25 . 

d.r" 

Here u = e" x , Mj = oe", ••• u,^ = a n ~ l e ax , u n = a n e ax . 
v = x, v l = l ) v. 2 = 0, v 3 = 0, •••. 

Substituting in (3), we have 

^ = — (e**x) = a n e ax x + m"" 1 ^ 1 =a tl - 1 e ax (ax + w). 

3. , = ( . + i )V ^ri; ^3(5^-14,4-13). 
d7? 8(»-l)* 

a -i d A y J, ,46,8 6\ 

1 ^ lo ^ J = € X l °Z X + x-^^-x*) 

K ,31 /o , in c? 4 V 48(a;4-l)(2« 2 + 2x + l) 

6. 2/ = sin x log cos a,*, — - = sin a; [log cos x — 2 tan 2 #(3 tan 2 x -*- 5)1. 

dar 

7. y = x*a*, ^ = a x (loga)"- 2 [(ajloga + ft) 2 -ft]. 

\ 

8. v = ! > — -=( — 1) ft- * — ! ! — • 

9 (a-,+ 1) 3 dx n K } - (x + l) n+ * 



CHAPTER V 
DIFFERENTIALS. INFINITESIMALS 

61. The derivative -^ has been defined, not as a fraction having a 

ax 

numerator and denominator, but as a single symbol representing the 
limiting value of — , as Ax approaches zero. In other words, the 

derivative has not been defined as a ratio, but as the limit of a ratio. 

We have seen (Art. 56) that derivatives have certain properties 

of fractions, and there are some advantages in treating them as such, 

thus regarding -^ as the ratio between dy and dx. 
dx 

Various definitions have been given for dx and dy, but however 
defined, they are called differentials of x and y respectively. The 
symbol d before any quantity is read " differential of." 

62. Definition of Differential. One definition is the following: 
The differential of any variable quantity is an infinitely small in- 
crement in that quantity. That is, dx is an infinitely small Ax, and 
dy an infinitely small Ay. 

By the direct process (Art. 16) of finding the derivative of an 
algebraic function, Ay is generally expressed in a series of ascending 
powers of Ax, beginning with the first. 

For example, if y = x 3 , y + Ay=(x + Ax) 3 , 

and Ay = 3x 2 Ax + 3x(Axy+(Ax)\ ... (1) 

In finding the derivative we have 

^ = 3x? + 3xAx + (Ax) 2 , 

in which, as Ax approaches zero, the second member approaches 3x? 
as its limit, the second and third terms approaching the limit zero. 






DIFFERENTIALS 69 

If we let A.r approach zero in equation (1), every term approaches 
zero, but there is nevertheless a marked distinction between them, in 
that the second and third terms, containing powers of Ax higher than 
the first, diminish more rapidly than that term. 

Thus we have Ay=3x 2 Ax approximately, 

and the closeness of the approximation increases as Ax approaches 
zero. 

From this point of view, regarding dx and dy as infinitely small 
increments, we may write 

dy = 3x 2 dx, 

not in the sense that both sides ultimately vanish, but in the sense 
that the ratio of the two sides approaches unity. 

Thus dy=3x 2 dx, and ^- = 3x 2 , 

dx 

are two modes of expressing the same relation. 

According to the first, 

An infinitely small increment of y is 3x 2 times the corresponding infi- 
nitely small increment of x. 

According to the second, 

Tlie limit of the ratio of the increment of y to that of x, as the latter 
increment approaches zero, is 3xr. 

Just as we sometimes say 

"An infinitely small arc is equal to its chord," instead of 
"The limit of the ratio between an arc and its chord, as these 
quantities approach zero, is unity." 

So in general, if y =f(x), 

Lim Aae=0 -^=/'(aj), 

Ax 



that is, ^/ =/'(*) + e , 

Ax 

where c approaches zero as Ax approaches zero. 



70 



DIFFERENTIAL CALCULUS 



Hence 



Ay = f'(x)Ax-\-eAx, 



and as the term eAx diminishes more rapidly than the term f'(x)Ax } 
we have 

Ay = / '(a?) Ax approxim ately , 

or dy=f'(x)dx. 

Corresponding to every equation involving differentials, there is 
another equation involving derivatives expressing the same relation, 
and the former may be used as a convenient substitute for the more 
rigorous statement of the latter. 

Thus the use of differentials is not indispensable, but convenient. 
It should always be kept in mind that their ratio only is important, 
the derivative being the real subject of mathematical reasoning. 

63. Another Definition of Differentials. The differentials dy, dx, 
are sometimes defined as any two quantities whose ratio equals the 

derivative 



dy 
dx 



tan RPT. 



Y 














o/ 






?^<k\ 


T 
R 




y 


dx 










X 









Let us see what this defini- 
tion means geometrically. 

If we regard the derivative 
as the slope of a curve, 
dy 
dx 

By this definition of differen- 
tials, dx may be any distance 
PR taken as the increment of 
x, and dy is then RT, the corre- 
sponding increment of the ordi- 
nate of the tangent line at P. 

That the two definitions are 
consistent will appear, if we 
diminished. 

The smaller we take PR, the more nearly is ^- equal to unity, or 

RQ 
in other words, the more nearly is RT equal to RQ. 

If PR is supposed to be infinitely small, this definition of differ- 
entials becomes that of the preceding article. 



suppose PR to be indefinitely 
RT 



DIFFERENTIALS 



71 



The second may be said to be the more rigorous of the two defini- 
tions, but the first has the advantage of being more symmetrical, and 
better adapted to the various applications of the calculus to mechanics 
and physics. 

64. Formulae for Differentials. The formulae for differentiation 
may be expressed in the form of differentials by omitting dx in each 
member. 

To each of the formulae for a derivative, corresponds a formula 
for a differential. 

Thus we have 



II. 


dc = 0. 


III. 


d(u + v) = du + dv. 


IV. 


d(uc) —vdu -f udv. 


VI. 


jfu\ _ vdu — udv 


VII. 


d(u") = nu n ~ l du. 


IX. 


7 1 (l " 

d log u = — 

u. 


XI. 


de" = e , 'du. 


XIII. 


d sinw = cosu du. 


XIV. 


d cos u = — sin u du. 


XV. 


d tan u = sec 2 udu. 


XVI. 


d cot u= — cosec 2 u du. 


XVII. 


d sec u = sec u tan u du. 


XVIII. 


d cosec u = — cosec u cot u du. 


XX. 


7 • _i fl " 


O Sill U — • 




Vl-u 2 


XXII. 


7 *. -1 (l " 

,i tan u = -• 

1 +U* 


XXIV. 


i du 

d sec u = 

"Vu Y ^l 


XXVI. 


7 , du 

il vers - u = • 



V2u 



72 DIFFERENTIAL CALCULUS 

Differentiation by the new formulae is substantially the same as 
by the old, differing only in using the symbol d instead of — . 

(XX 

For example, let y = 2 . 

X -\- O 

to, _ rif^±l\ - (x 2 +3)d(x + 3)-(x + 3)d(x 2 + 3) 
ay ~ a [x^T3j~ (z 2 + 3) 2 

= (x 2 + 3) dx - (x + 3) 2 xdx 

~ (x 2 + 3) 2 

_ (a; 2 + 3 - 2 a; 2 -6 apcfa = (3 - 6 a; - a 2 ) cfa 
(x 2 + 3) 2 (a; 2 + 3) 2 

If we wish to express the result as the derivative, we have only 
to divide by dx, giving 

dy_ 3 — 6x — x 2 

dx~ (x 2 -\-3) 2 



EXAMPLES 

Differentiate the following functions, using differentials in the 
process : 

1. y=(a?-l)(2-3a>)(2a> + 3), dy= (-IS x 2 + 2 x + ll)dx. 

2 x = (t-l)(t-2) dx= 6(f-2)dt 

(* + l)(* + 2)' (t+l)«(* + 2)«" 

3. y = -vVTT Va7^2, dy = ~Z= rdx- 

Wx'-\- 1 Var — 2 

4. r = 5J2l-*, ^ = (2 + S m^)sin^ 

cos 3 cos 4 

5. y = e*(x 3 -6x 2 + 2£x-4:0), dy= e 2 (^ + Adx. 



INFINITESIMALS 73 

f6. r = sin 6 log tan 9, dr = cos 6 log tan 6 dO + sec 6 dO. 



4.r , 4 cfa 

7. w = tan * -, o?/= . 

4 - sc 8 5 4 + x 2 

sin- 1 3a + 3.r Vl - 9 x 2 , dy = S VI — 9 x* dx. 

3 tan 2 (9 dO 



9. <^ = tan-Han 3 ^, d^ = 



tan 4 6> — tair^+1' 



65. Order of Infinitesimals. In Art. 62 we have spoken of infinitely 
small or infinitesimal increments. 

An infinitesimal may be defined as a variable whose limit is zero. 

If there are several infinitesimals that approach zero simulta- 
neously, one of them, a, may be taken as the standard of comparison 
and called the principal infinitesimal. 

Then a 2 , a 3 , a* are said to be infinitesimals of the second, third, 
?ith orders, with respect to a. 

In general the order of an infinitesimal is defined as follows: An 
infinitesimal (3 is said to be of the nth order with respect to a when 

Lim a=0 ^ = A', a finite quantity, not zero. . . . (1) 

When n= 1, (3 is of the first order with respect to a. 
When n = 2, (3 is of the second order with respect to a. 

From the definition it may be shown that the limit of the ratio of 
an infinitesimal to one of the same order is finite, and to one of a 
lower order, zero. 

Equation (1), Art. 62, illustrates infinitesimals of different orders. 
If we write it 

dy = 3x 2 dx + 3x (dx) 2 + (dx) 8 , 

and regard dx as the principal infinitesimal, the terms of the second 
member are infinitesimals of the first, second, and third orders, with 
respect to dx. 

i), if we regard x as the principal infinitesimal, of what orders 
are Bin a; and \tvsx, with respect to x? 



74 DIFFERENTIAL CALCULUS 



By Art. 12, Limx=o ?HL£! = l ? a finite quantity. 

Hence by (1) sin x is an infinitesimal of the first order with respect 
to x. 

T . versa; T . 1 — cosx siii 2 sc T - 1 /sinscV 1 

Lim x=0 — — = Lim x=0 — — — ■ — — == Lim x=0 



x- sin' 2 * or l + cosa?y x J I 

a finite quality. 

Hence by (1) versa? is an infinitesimal of the second order with 
respect to x. 

Show that tan 6 — sin 6 is an infinitesimal of the third order with 
respect to 6. 



CHAPTER VI 
IMPLICIT FUNCTIONS 

(See also Art. 114.) 

66. In the preceding chapters differentiation has been applied to 
explicit functions of a variable. The same rules or formulae of differ- 
entiation are sufficient for deriving -^, —4, — *■, •••, when y is an 

° dx dx 2 ' dx 5 ' ' y 

implicit function of x ; that is, when the relation between y and x is 
expressed by an equation containing these variables, but not solved 
with respect to y. 

For example, suppose the relation between y and x to be given by 
the equation 

9 9,799 97 9 

cry 2 + b 2 xr — orb'. 
Differentiating with respect to x, 

dx 



2a 


h J ( hl + 2b i x 
dx 


= o, 


dx 


b 2 x^ 
a 2 y 





Having thus obtained the first derivative, we may by another 
differentiation find the second derivative. 



c C-ylf- - Hm»3K b 2 fy -x&S 
J dx V dx) 



dx- d.ca 2 y~ <i^f d 2 y 2 

7-3 



76 DIFFERENTIAL CALCULUS 



Substituting now for -^ its value, 
dx 



dry _ 


b\a 2 f 


+ b 2 x 2 ) 


b* 


dx 2 


a 


y 


ay 


differentiatin 


g again, 


we may 


obtain 


d 3 y _ 
dx 8 


3 6\ 
a 4 y 5 ' 







The first differentiation may be conveniently performed by differ- 
entials instead of derivatives. Thus we should have from the 
equation 

a 2 y 2 -f b 2 x 2 =a 2 b 2 , 

2a 2 ydy + 2b 2 xdx = 0, 

dn "hi* 

giving -^ = — , as before. 

dx a 2 y 

In deriving — ^, — *, ..., derivatives should be used rather than 
differentials. dx dx 



EXAMPLES 

Find the following derivatives. 
1. (x- a ) 2 + (y-b) 2 = c 2 , 

''_ d 3 y_ 3c 2 (x-a) 



d v _ 


x — a d 2 y _ 


dx 


y — b' dx 2 




dy_xy- y 2 




dx xy + x 2 



(y-by dx* (y-by 

2. x=y\og(xy), 

3. (coB^=(8in^, gj = logBin* + <frtanfl 

7 v ry ' dB Iogcos0-Ocotcf> 

4. ax 2 +2 hxy-\-by 2 = 1, 

<fy_ _axjj-hy d 2 y _ lr — ab d 2 x _ h 2 —ab 
dx hx + by' dx 2 (hx + by) 3 ' dy 2 (ax + hyy 



IMPLICIT FUNCTIONS 77 



5. aa? + 2hx!r + bf = } d / = «. 

dx x 

. . , . rtd> sin 2 6 d 2 d> 2 sin 2 Afcos 2 <f> + cos 2 0) 

6. tan tand> = w,— " — : — _x _x = ^ v . T* z - 

r 'd0 sin2 0'c?0 2 sin 2 2 5 

„ o / \ i / s ch/ 2)/ — x d 2 )/ (x — y) 2 

8. (3x + y + 6y(3y-3x + 2)=c, Jj=y=|e. 

9. ,W« + 2r + l = 0, (|J + 2,eot.|=^ 

10. e-=crtr diJ = ?/ "~ logft , ^y = %-iog«) j 

c?£ a; — log 6 ' (to 2 (a; — log b) 2 ' 

hi / o , o N o . i v r?y « + ?/ d 2 y 2x? + 2w 2 

ii. log (,-+,-) = 2 tan -^, g=^-J y , J=-^r#- 



CHAPTER VII 
SERIES. POWER SERIES 

67. Convergent and Divergent Series. The series 

%+w 2 -f u 8 + ••• +u n + u n+1 , . . ... . . (1) 

composed of an indefinite number of terms following each other 
according to some law, is said to be convergent when the sum of the 
terms approaches a finite limit, as the number of terms is indefi- 
nitely increased. But when this sum does not approach a finite 
limit, the series is divergent. That is, if S n denote the sum of the 
first n terms of (1), the series is convergent, when 

Lim )(=00 S n = some definite finite quantity. 

When this condition is not satisfied, the series is divergent. 

Thus the geometrical series, 

a + ar +ar 2 -f- ar 3 -f- 

is convergent when r is numerically less than unity, and divergent 
when r is numerically greater than unity. 

For S n = a + ar + ar* + ••• + ar"" 1 = a (} ~ **) . 

1 — r 

When * \r | < 1, Lim^ S n = ^_i_ . 

1 — r 

When | r | > 1, Liin n=00 S n = oo. 

When | r | =1, the series is also divergent. 

68. Series of Positive and Negative Terms. Absolute and Conditional 
Convergence. In the case of series composed of both positive and 
negative terms, a distinction is made between absolute convergence 
and conditional convergence. 

* | r | denotes the numerical value of r. 
78 



SERIES 79 

Before defining these terms, the following theorem should be 
noticed: 

A series whose terms have different signs is convergent if the series 
formed by tailing the absolute values of the terms of the given series is 
con vergent. 

"Without giving a rigorous proof of the theorem, we may regard 
the given series as the difference between two series formed of the 
positive ami negative terms respectively. 

The theorem is then equivalent to this : 

If the sum of two series is convergent, their difference is also 
convergent. 

A series is said to be absolutely convergent, when the series of the 
absolute values of its terms is convergent. 

A series whose terms have different signs may be convergent 
without being absolutely convergent. Such a series is said to be 
conditionally convergent. 

For example : 1 1 h (1) 

converges to the limit log e 2, 

but it is not absolutely convergent, since 

2 3 4 
is divergent (see Art. 70). 

Series (1) is accordingly conditionally convergent. 

But l_i + I_I + ... 

2 2 3 2 4 2 

is absolutely convergent (see Art. 70). 

69. Tests for Convergence. The following are some of the most 
useful tests. 

In every convergent series the nth term must approach zero as a 
limit, as n is indefinitely increased. 

That is, the series v^ -f- u 2 + Mg+ ••• + u n -\ 

is convergent, only when Lim # , =x u n — 0. 

For S n = #__, + u n . 



80 DIFFERENTIAL CALCULUS 

If the sum of the series has a definite limit, 

Lim n = a0 S n = Lim n=0O S n . v 
Hence Lim n=00 M n = (1) 

For a decreasing series whose terms are alternately positive and 
negative, this condition is sufficient.* 

For example, 1 — - + = + -••* 

^ ' 2 3 4 

is convergent. But the decreasing series 

2 _ 3 4 _ 5 
1 2 3 4*" 

is divergent, as it does not satisfy (1), since Lim M=00 u n = 1. 

The sum of this series oscillates between two limits, log e 2 and 
1 + log e 2, according as the number of terms is even or odd. Such a 
series is called an oscillating series. 

For a series whose terms have the same sign, the condition (1) is 
not sufficient. For example, the harmonic series 

^2^3 4^ 
is divergent (see Art. 70). 

70. Comparison Test. We may often determine whether a given 
series of positive terms is convergent or divergent, by comparing its 
terms with those of another series known to be convergent or diver- 
gent. 

In this way the harmonic series 

1 + - + - + -+- + - + - + - + — . • . • . (1) 

2 3 4 5 6 7 8 w 

may be shown to be divergent, by comparing it with 

1+ I + H + H + H + - (2) 

* The proof of this is omitted. 



SERIES 81 

Each term of (1) is equal to, or greater than, the corresponding 
term of (2). Hence if (2) is divergent, (1) is also divergent. But 
(2) may be written 

T 2 T 4816 
= 1 + 1 + 1 + 1 + 1 + ... 

The sum of this series is unlimited ; hence (2) is divergent, and 
therefore (1). 

Consider now the more general series 

h+h + h + h+- (3) 

If p = 1, the series (3) becomes (1), which is divergent. 
If p < 1, every term of (3) after the first is greater than the cor- 
responding term of (1). Hence (3) is divergent in this case also. 
If p > 1, compare 

! + !+!+L + L + !+!+L + ...+JL + ... . (4) 

1» 2' 3' 1' o» 6' 7' 8' 15" v i 

™ th h + h + h+h + h + h + h+h +, " + h + " : - • ® 

Every term of (4) is equal to, or less than, the corresponding 
term of (5). But (5) may be written 

L + ^+l+^ + ... 

\P ' 9 P 4: P S p ' 

2 

a geometrical series whose ratio, — , is less than unity. 

Hence by Art. 67, (5) is convergent and consequently (4). 
Thus it has been shown that 

whenp^l, the series (3) is divergent; 
when p >1, the series (3) is convergent. 

The series (3) together with the geometrical series are standard 
series, with which others may often be compared. 



82 



DIFFERENTIAL CALCULUS 



71. Cauchy's Ratio Test. This depends upon the ratio of any 
term to the preceding term. In the series 



u 2 + u 3 H h u n + u n+1 + • • 



(1) 



this ratio is 



Let us first consider, from this point of view, the geometrical 

.... (2) 



series 



a + ar + ar -\ \- ar n 4- ar n+1 + 



Here the ratio -^ = r, and is the same for any two adjacent terms. 

We have seen (Art. 67) that this series is convergent or divergent, 

according as , , ■ „ , , i - 

|r|< 1, or |?-|> 1. 

That is, (2) is convergent or divergent according as 



< 1, or 



>1. 



If now (1) is any series other than the geometrical series, the 
is not constant, but a function of n. The series is then 



ratio 



Wi 



convergent or divergent, according as 



Liim 



< 1, or Lim, J= 



Wi 



>1. 



(3) 



We will first suppose (1) to be a series of positive terms. 



Let 



Lim, 



= P . 



Suppose p < 1. By taking n sufficiently large we can make 
— approach its limit p as nearly as we please. 

n 

There must be some value m, of n, such that when n ^ m, 



-^- < r, a proper fraction. 

Hence u m+1 < ujr 3 u m+2 < u m+l r < u m r 2 , etc. 

u m + u m+l + m ot+2 +-- <u m + u m r+ uy+ ••-. . 



• W 



SERIES 83 

But since r < 1, the second member of (4), which is the geometrical 
series, is convergent, and therefore the first member 

is convergent. Consequently (1) is convergent. 

Suppose p > 1. By similar reasoning, when n > m, 

-^ > r, an improper fraction. 

Hence v. m ^ > u m r, u m+2 > u m+1 r > u n f+ etc. 

Since r > 1, the second member, and therefore the first member, 
must be divergent. 

Thus the theorem is proved for a series of positive terms. 

If the terms of (1) have different signs, it is evident from Art. 68 
that the series will be absolutely convergent if 

u. 



Lim, 



l n+l 



<1. 



It is also true that for different signs, (1) will be divergent if 

,n+l 



Lira, 



>1. 



If Lira, 



The proof of this latter statement is omitted. 

U n+1 _ -i 

the series may be either convergent or divergent. There are other 
tests for such cases, but they will not be considered here. 

EXAMPLES 
1. Is the following series convergent? 

1-2 2- 2 2 3.2 s «2» 



Applying (3), Art. 71, we have ^*±* 



n 



u n 2(w + l) 

1 



2 (n + 1) 2 
As this is less than unity, the given series is convergent. 
Its limit is log, 2, as will appear later. 



84 



DIFFERENTIAL CALCULUS 



Determine which of the following series are convergent, and which 
divergent. 



2 ' 2 + \3 + g + \5 + 



+ 



II 



+ 



10 10 2 10 3 10 4 



4 - l+ l + H + 



5 1 1 1 I 1 1 1 I 



6. l-f? + l ± ^ 
2 2 2 2 3 2 4 



7 . 1+ '* + * * 

L? L§ Ji= 



8 i_?+?-^+A_ 

3 5 7 9 11 



9 1+I+A.+ , i 

' 2 _r 5" r 10 n 2 + l 



+ 



10. 



1 4— i-_ + 1 



1 + Vl 1 + V2 1 + V3 



11. log?-log| + log|-log| + 



By (3), Art. 71. 

By (3), Art. 71. 
By (1), Art. 70. 



By (1), Art. 69. 

Compare with (3), Art. 70. 

Compare with (3), Art. 70. 

By (1), Art. 69.1 



12. sec^-see^+sec^-sec^ + 



13. sin 2 ^ + sin 2 - + sin 2 ^ + sin 2 ^ + .. 
2 3 4 5 



POWER SERIES 



85 



14. 



15. 



16. 



1 + 1 

1 2 + 1 2 2 + l ' 3 2 + l 4 2 + l 

1 + 1 2 + 1 3 + 1 4 + 1 
l» + l~ t "2 2 +l~ t S 2 + l~ f "4 s + l 



2+1 , 3 + 1 4+1 , 

' OS . i .19 -t ' 



+ 



2+l + 3±l + 4 ± l + 5 ± l + 



2^-1 33_1 4 3_1 5 S_ ± 

Answers 
Exs. 2, 5, 6, 9, 11, 13, 14, 16, convergent. 
Exs. 3, 4, 7, 8, 10, 12, 15, divergent. 
Exs. 8, 12, oscillating. 

72. Power Series. A series of terms containing the positive in- 
tegral powers of a variable x, arranged in ascending order, as 

a + a x x + a 2 # 2 + a$? -\ , 

is called a power series in x. The quantities a,-,, a l5 a 2 , 
supposed to be independent of x. 

For example, l + 2a; + 3a,- 2 + 4^ 3 H , 



are 



1_ t+ I*_ t 

12 li 1$ 

are power series in x and y respectively. 



+ 



73. Convergence of Power Series. A power series is generally 
convergent for certain values of the variable and divergent for 
others. 

If we apply the ratio test, (3), Art. 71, to the power series 

a {) +a x x + a#?-\ \-a n x n -\ , (1) 

we have for the ratio between two terms 



Un±X 



a n x 



Lim. 



= Lim, 



a n x 



= |a?|Liin BS 



«n-l 



86 DIFFERENTIAL CALCULUS 

The series (1) is convergent or divergent according as 



| x | Li m^ 

that is, according as 

|<c|<Lim n= 

The case \x\ = Lim w=Q 



< 1, or | a; | Lim, 



>i; 



or |#|>Lim 7t= 
requires further examination. 



For example, consider the series 

1 + 2x + 3x 2 + 4lX 3 -\ t-naT- 1 + (n + 1) a? +—. 

Here = , Lim,,^, = 1. 



a n n + 1 



71 + 1 



(2) 



Hence (2) is convergent or divergent, according as 
|aj| < 1 or \x\ > 1. 

We may say that (2) is convergent when — 1 < x < 1, and the in- 
terval from — 1 to + 1 is called the interval of convergence. 

EXAMPLES 

Determine the values of the variable for which the following 



1. l + x + x 2 + x s +--- 



x , x- 



2. -J^-j-_^_ + -^ + 
1-2 2-3 3-4 



/v*— /y»3 rW* 

3. »+'-+- + - + —. 
,2^3 4 

A X s , X 5 x 7 . 

4 - X -3 + 5~7 + -' 



POWER SERIES 87 



- .1 .r* , 1 ■ 3 .i- 3 , 1 • 3 ■ T> .r 7 , 

5- .r + ^ • - + — -T ~ + n , ,. - + • 



7.- m3 />v* 



V- )•"* }•' 

7. 1-— + — -=-+■•'. : ««x 

2 i [6 



Q a- 3 . sc 6 a*' . 

12 • 12 II 

Exs. 1-5, convergent when — 1 < a? < 1 
Exs. 6-8, convergent for all values of jc. 



CHAPTER VIII 
EXPANSION OF FUNCTIONS 

74. When by any process a given function of a variable is 
expressed as a power series in that variable, the function is said to 
be expanded into such series. 

Thus by ordinary division 

= l_ a; + ^_ x 3 + ... (1) 



1 + x 



By the Binomial Theorem 

(x + a) 4 = a 4 + 4 a 3 x+ 6 a¥ + 4 ax 3 + x*. 

(l-£ C )- 2 = l + 2x + 3» 2 + 4ar 3 +... (2) 

The methods employed in these expansions are applicable only to 
functions of a certain kind. We are now about to consider a more 
general method of expansion, of which the foregoing are only special 
cases. 

It should be noticed that when a function is expanded into a 
power series of an unlimited number of terms, as (1) and (2), the 
expansion is valid only for values of x that make the series con- 
vergent. For such values, the limit of the sum of the series is the 
given function, to which we can approximate as closely as we please 
by taking a sufficient number of terms. 

The general method of expansion is known as Taylor's Theorem 
and as Maclaurin's TJieorem. 

These two theorems are so connected that either may be regarded 
as involving the other. We shall first consider Maclaurin's 
Theorem. 

88 



EXPANSION OF FUNCTIONS 89 

75. Maclaurin's Theorem. This is a theorem by which a function 
of x may be expanded into a power series in x. It may be 
expressed as follows : 

/(*) =/(0) +/(Q)j+/' (0)|+/"(0)|+ -, 

in which f(x) is the given function to be expanded, and/' (x),f (x), 
f"(x), •••, its successive derivatives. 

f(O),f(0),f'(0), • ••, as the notation implies, denote the values of 

/(*), / (*)> /' (*)> ' ' ; when * = °- 

76. Derivation of Maclaurin's Theorem. If we assume the 
possibility of the expansion of f(x) into a power series in x, we may 
determine the series in the following manner: 

Assume 

/(x^i + m'+CV + DxH^i •••, - . . . (1) 

where A, B, C, ••• are supposed to be constant coefficients. 

Differentiating successively, and using the notation just defined, 
we have 

f\x) = B + 2Cx + 3Dx 2 + 4:Ex i + ..- .... (2) 

f"(x) = 2C + 2.3Dx + 3.±Ex 2 ... (3) 

f"'(x) = 2.3D + 2-3-4;Ex + ... . (4) 

r(x) = 2'Z-±E+... (5) 



Now since equation (1), and consequently (2), (3), ••• are supposed 
true for all values of x, they will be true when x = 0. Substituting 
zero for x in these equations, we have 



from (1), f(0)=A, A=f(0), 

from (2), f(0)=B, B=f(0), 

from (3), /'(0) = 2C, C =^' 



90 DIFFERENTIAL CALCULUS 



from (4), /»'(0) =2- 3D, D = OS, 

from (5), / iv (0) = 2-34^ E = G^-, 

li 



Substituting these values of A, B, C, • • • in (1), we have 

/(*)=/(0)+/ , (0)j+/"(0) | | + /"'(0)^+--- ... (6) 

77. As an example in the application of Maclaurin's Theorem, let 
it be required to expand log (1 -f x) into a power series in x. 

fix) = log (1 + x), /(0) = log 1 = 0. 

f(x) = -±- = (l + x)-\ /'(0) = 1. 

f'(x)=-(l + x)-*, /"(0)=-l. 

/'»(*)= 2 (1+z)- 3 , /'"(0)=2. 

/* (x) = - [3 (1 + *)- 4 , /"(0) = - [3. 

/*.(*) = 14(1+.*)^, /- (0) =14. 



Substituting in (6), Art. 76, we have 

x 2 , 2 x s [3* 4 , Li* 5 
log(l+x) =0 + 1.^-1- 2 +^--^J + ^g---- 

/v?2 /yiO /yi4 /yi5 

iog(i-M)=*-f +!-!+!-•••. 

78. If in the application of Maclaurin's Theorem to a given 
function, any of the quantities, f(0), /'(0), /"(0), ••• are infinite, 
this function does not admit of expansion in the proposed power 
series in x. 

In this case f(x) or some of its derivatives are discontinuous for 
x = 0, and the conditions for Maclaurin's Theorem are not satisfied 
(see Art. 94). 

The functions logic, cotcc, x2 } illustrate this case. 



EXPANSION OF FUNCTIONS 91 



EXAM PLES 

Expand the following functions into power series by Maclaurin's 

Theorem : 

r- t 3 x 4 
1. e x = 1 + x + - : + '— 4- '— H • Convergent for all values of x. 



\2 |3 |4 



2*^ X° X' 

2 sin x = x — — + '— — ~ H • Convergent for all values of x. 

[3 [5 [7 



,2 ,4 6 

3 cos a; = 1 — — + - — + Convergent for all values of x. 

j2 |4 [6 

4. (a + jc)« = a" + va»-\v -f "^~^ a"~V 

+ "O*- 1 )!"- 2 ) ^-3^3 + ... Convergent when \x\ < a. 

[3 

5. log u (l + a;) = log ae ^-|+|-j+- 

Convergent when |a;|< 1. 

2 3 4 

6. log (1—*) = — x - — — — — — . Convergent when | x | < 1. 

7. tan- 1 ;/— x — — + — — ^--\ . Convergent when \x\ < 1. 

O O 4 

Here /(a?) = tan -1 a;, 

/'(*) =^-1-^ = 1 -o'+^-aM- ..., 



/" (a?) = -2a? + 4ar 3 - 6a? 5 + 



Q . . , 1 .r ' , 1 . 3 a? 8 1 • 3 • 5 x 7 . 



Convergent when \x\ <1. 



92 DIFFERENTIAL CALCULUS 

Here f(x) — sin -1 #, 

VI — XT 

Expanding by the Binomial Theorem, 

/' (a;) = 1 +aar>+ bx 4 + cx 6 + • ••, 

where a = -, 5 = —, c^ 1 ' 3 ' 5 , » 

2* 2-4' 2.4.6' 

/" (x) = 2 ax + 4&c 3 + 6ca 5 + — , 



9. sin (x + a) = sin a + a; cos a — — sin a — — cos a + 



Convergent for all values of x. 



10. log(l + x + x*) = x + ^-?f + ^ + ^ 

^ o 4 o 



Convergent when | x | < 1, 



. x 3 x 5 



11. e* sin x = cc + a3 2 H . Convergent for all values of x. 

12. e* cos # = 1 + a; 1 . Convergent for all values of x. 

3 6 

13. tana = a + ^ + ? r ?+"-. 

3 lo 

14. S ec* = l + | 2 + |f+-. 

15. logsec* = - + - + - + .... 

Defining the hyperbolic sine, cosine, and tangent by 

sinh x = — — — , cosh x = — — , tanh x = , show that 

2 ' 2 ' e-H-e-' 

16. sinh «=* + - + -+.-. 

[3 |5 



EXPANSION OF FUNCTIONS 



93 



ar , x" 



17. cosh 3 = 1 + ;- + ;-+. •■ 



x 3 2 r 5 
18. tanh x = x — — 4- 77- 
3 15 



19. Show by means of the expansions of Exs. 1, 2, 3, that 



,xiA-l 



cos as 4- V— 1 sin x, 



e -x v/- 1 _ cos a- _ -y/ _ 1 s i n ^ # 



These are important relations. 



79. Huyghens's Approximate Length of a Circular Arc. 

If s denote the length of the arc ACB, a its chord, and b the 

chord of half the arc, it may be shown that 

8 b - a . ., 

s = — - — , approximately. 
o 

Let <£ be the half angle AOC. 

Then s = 2 r<j>, and by Ex. 2, Art 72, 



a=2r sm<£ = 2W <f> 



*! + £ 

[3 [5 



in| = 2r^-^4--* 5 



2 V 2 2 ' 3 [3 25 1^ 

Combining so as to eliminate <£ 3 , 

b-a = 2rfs^ + ^=3s(l-^. 




480 



If s is an arc of 30°, & = —, and the error < - 

,T 12' 102000 



If s is an arc of G0°, <*> = |, and the error < ^q 



94 DIFFERENTIAL CALCULUS 

80. Computation by Series. 

Compute by Ex. 1, p. 91, Ve to 5 decimal places. 

Ans. 1.64872. 

Compute ^\/e to 10 decimal places. Ans. 1.1051709181. 

Compute by Ex. 2, p. 91, sin 1° to 8 decimal places. 

7T = 3.14159265. Ans. 0.01745241. 

Compute to 4 decimal places the cosine of the angle whose arc is 
equal to the radius. Ans. 0.5403. 

81. Calculation of Logarithms. By means of the expansion of 
log (1 + x), Art. 77, the Napierian logarithms of numbers may be 
computed. 

Let us find the logarithms in the following table. 

log 2 =0.6931, 

log 3 =1.0986, 

log 4 =1.3862, 

log 5 =1.6094, 

log 6 =1.7917, 

log 7 =1.9459, 

log 8 =2.0793, 

log 9 =2.1972, 
log 10 = 2.3025. 

It is only necessary to calculate directly the logarithms of the prime 
numbers 2, 3, 5, 7, as the others can be expressed in terms of these. 
We have from Art. 77, 

lo g 2 = log(l + l)=l-i + |-i+.... 

This series is convergent, but converges so slowly that 100 terms 
would give only two decimal places correctly. But we may obtain 
a series converging much more rapidly by taking 

log2 = log ? = log (1 + |)-log (1-|). 

~3 



EXPAXSIOX OF FUNCTIONS 



95 



For log 



1 + x 

1-x 



log (1 + X) - log (1 - x) 



.v- , x 3 a* , 

= X + 

2 3 4 






(-a 



ar _ ar _ .t 4 _ . 
2~ 3 4 } 

Convergent when !a;|<l. 



Thus log2 = 2/^ + -i- + -i- + -i-+...\ 

Four terms of this series give losr 2 = .6931. 



The computation may be arranged as follows: 



J- 

3 ~ 


.333333 


i =.333333 


1 
3 3 


.037037 


— = .012346 
3-3 3 


1 _ 
3"' 


.004115 


1 =.000823 
5-3 5 


1 
3 7 


.000457 


1 . = .000065 


1 
3 ,J 


.000051 


1 r = .000006 
9 • 3" 




.34657 






2 



.69314 
The numbers in the first column may be obtained by dividing suc- 



cessively by 9. 



1 + x 



1 + 



Any number may be put in the form T an( j \ Q(T 3 



may be found like log 2. 



DIFFERENTIAL CALCULUS 



But having log 2, it is easier to compute 
1 



1 + 



log - = log 



log3 = log| + log2. 



and then 

Let the student make this computation 

Find log 5 from 



i+i 



i . 5 ■■! '4 

log - = log - 



1- 



In a similar way find log 7 from log 5. 

Having obtained the logarithms of 2, 3, 5, 7, find the other loga- 
rithms in the table at the beginning of this article. 

To obtain the common logarithm, that is, logarithm 10 , it is only- 
necessary to multiply the Napierian logarithm by .4343, the modulus 
of the common system. 

Find thus the common logarithms of the numbers in the foregoing 
tables, — first, of 2, 3, 5, 7, and from these the others. 

82. Computation of ir. From Ex. 7, p. 91, by letting x = 1, we have 

I = taa->l = l-! + H + .., 

a slowly converging series. 

To obtain a series converging more rapidly, we may use 
tan" 1 1 = tan" 1 - + tan - 1 



3' 



from which 



1 + 1 



1 



4 2 3 • 2 s ' 5 • 2 5 7 • 2 7 



+ 



^3 3-3 3 5-3 5 7-3 7 



+ 



EXPANSION OF FUNCTIONS 97 

By taking 9 terms of the first series and 5 of the second, the 
student will find 

- = 0.463647 ... + 0.321751... 
4 

and 7T = 3.14159 ••.. 

Other forms of tan -1 1 may be used, giving series converging even 
more rapidly, as 

tan" 1 1=2 tan" 1 - + tan" 1 - • 



tan -1 1 = 4 tan -1 tan-i 

5 239 

By these formulae the computation has been carried to 200 deci- 
mal places. 

83. Taylors Theorem. This is a theorem for expanding a function 
of the sum of two quantities into a power series in one of these 
quantities. 

As the Binomial Theorem expands (x + h) n into a power series 
in h, so Taylors Theorem expands f(x + h) into such a series. It 
may be expressed as follows : 

/(x + ft)=/(x)+/'(x)A+/"(x)|+/"'(»)|+-. 

84. The proof of Taylor's Theorem depends upon the following 
principle : 

If we differentiate f(x + h) with respect to x, regarding h con- 
stant, the result is the same as if we differentiate it with respect 
to h, regarding x constant. 

That is, -ff(x + h) =-£ f(x + h). 

ax all 

For, let z = x + h, (1) 



98 DIFFERENTIAL CALCULUS 

then by (3), Art. 56, 

dx dx dz dx 

dh Jy J dJi K } dz K J dh 

But from (1), — = 1. and — = 1 ; 
v ; ' dx ' dh 

therefore Af( x + h)= — f (x '+ h) . 

dx JK ^ J dh JKJ 

85. Derivation of Taylor's Theorem. If we assume the possibility 
of the expansion of f(x + h) into a power series in h, we may deter- 
mine the series by the aid of the preceding article. Assume 

f(x+h) = A + Bh + Ch 2 + m s + (1) 

where A, B, C, • • • are supposed to be functions of x but not of h. 
Differentiating (1), first with respect to x, then with respect to h, 

d j., , 7N dA . dB 7 . dC.o , dD JS , 
dx dx, dx dx dx 

— f(x + h)=B + 2 Ch -f 3Dh 2 + .... 

By Art. 84, the first members of these two equations are equal to- 
each other, therefore 

§A + QRn + Mltf +... = B + 2Ch + 3Dh 2 + .... 
dx dx dx 

Equating the coefficients of like powers of h according to the 
principle of Undetermined Coefficients, we have 

^ = B, B = ^. 

dx dx 

dB_2f l p _ 1 d 2 A 

dx 2 dx 2 

dC=3D, j) = ~ c ~ 

dx [3 dx 3 



EXPANSION OF FUNCTIONS 99 

The coefficient A may be found from (1) by putting h = 0, as that 
equation is supposed true for all values of h. 

Then A =/(«). 

Hence £=^ =/'(»)• 



Substituting these expressions for A, B, C, ••• in (1), we have 

/(.r + /o=/«>o+/'(^A+rw|Vr'(^,4 3 +---. . . (2) 

[2 [3 

86. Maclaurin's Theorem may be obtained from Taylor's Theorem 
by substituting x = 0. We then have 

/CO =/(0) +./(0)» +/"(0)jf +/"'(0)jf + •- 

This is Maclaurin's Theorem expressed in terms of h instead of x. 

87. As an example in the application of Taylor's Theorem, let it 
be required to expand sin (x + h) into a power series in h. 

f(x + h) = sin (x + h) ; 
hence / (x) = sin x, 

f'(x) = cos x, 
f"(x) = — sin x, 
/"'(a*) = — cos x, 
/ iv (a;) = sin x. 



Substituting these expressions in (2), Art. 85, we find 

/r h* h 4 

sin (x + /ij = sin x -f h cos x — — sin a: — — cos x + — sm a: -f 

re. 



100 



DIFFERENTIAL CALCULUS 



EXAM PLES 

Derive the following expansions by Taylor's Theorem : 

h 2 h 3 

1. cos (x + h) = cos x — h sin x — 7F> C0S x + 75 s i n ^ + 

3. (x + /i) 7 = x 7 + 7a; 6 7i+ •••. 



4. (x + 7i) w = af 1 + nx n ~ x h + 



n (n— 1) 



+ 



A 4 



h h 2 h 3 

5. log (a; + h) = log x -\ 7Tl> + ^Js~T^4 + ' 

7 a? 2 ar 3 ar 4 ar 

6. tan (a; + h) = tan a; + h sec 2 a; 4- 7r sec 2 a; tan x 
+ ^(3sec 4 aj-2sec 2 x) + .... 

7. Compute from Ex. 1, cos 62° = 0.4695. 

8. Compute from Ex. 6, tan 44° = 0.9657, tan 46° = 1.0355. 
9 fjx + h) +/(»,- h) =f(x) + | r(x)+ V /iv(a;) + _ 

10 . /(«+*) -ffr-Q ^hfim) + g /"'(g) + g/» + -; 

As a special case of Ex. 10, derive 

ii og .«±» - * + j!. + i. + .., 

2 # — 7i a; oar 5X 5 

11. /(2a.) = /(a ; )+x/^) + ^ 2 /' , W + |r'W + -... 



12. / 



1 + aJ 



/{«) 



/'(*) + 



f" (x) 



1 + 05 



13. If 2/= /(#), show that 



(1 + *) 2 1 2 

x 3 f'"(x) 
(1 + *) 3 [3 



+ 



y da; ^cto 2 [2 ^da: 3 [3 



EXrAXSIOX OF FUNCTIONS 



101 



88. In the preceding derivations of Taylor's and Maclaurin's 

Theorems, the possibility of the expansion in the proposed form has 
been assumed. In the remainder of this chapter we shall show how 
Taylor's Theorem may be derived without such assumption. 



89. Rolle's Theorem. If a given function <£(.r) is zero when 
x = a and when x=b, and is continuous between those values, as 
well as its derivative <£'(.r); then 
<f>\x) must be zero for some value 
of x between a and b. 

Let the function be represented 
by the curve y = <f>(x). Let 
6 A = a, OB = b. Then accord- 
ing to the hypothesis,, y = when 
x = a, and when x = b. 



O/A B\ 

Since the curve is continuous 

between A and B. there must be 

some point P between them, where the tangent is parallel to 

and consequently <f>'(x) = 0. 



OX, 



90. Mean Value Theorem. If/(#) is continuous from x = a to 
x = b, there must be some value x x of x, for which 



b-a J K J 



This may be stated geometrically 
thus: 

The difference of the ordinates of 
two points of a continuous curve, 
divided by the corresponding differ- 
ence of abscissas of these points, equals 
the slope of the curve at some inter- 
mediate point. 

In the figure let the curve PRQ 
represent y=f(x). 







A 


1 


A 


( 




P /Y 












M 









K B 



102 DIFFERENTIAL CALCULUS 

Let OA = a, OB = b. Then 

b-a PM 

At some point of the curve, as R, between P and Q, a tangent can 
be drawn parallel to PQ. Call OK=x 1 . Then the slope of the 
tangent at R is/'(a^), which equals tan QPM. 

Hence H S~f( a ) =f( Xl ), where a < x, < b. . (1) 

— CL 

If we let AB = b — a = h, b = a + h, (1) may be written 

f(a + h)=f(a)+hf(a + cf>h), where 0<<£<1. . (2) 

91. Another Proof. The following method of deriving (2), Art. 84, 
is important, in that it may be extended to higher derivations of 
f(x), as appears in Arts. 92, 93. 

Let R be defined by 

f(a + h)-f(a)-hR = (1) 

That is, let R denote /( a + ?0 -/(«) . 

h 

Consider a function of x whose expression is the same as (1) with 
x substituted for h. Call this function 4>(x). 

That is, <£(#) =f(a + x) -f(a) -xR (2) 

Differentiating, <j>' (x) = f (a + x) - R (3) 

It is evident from (2) that <j>(x) = 0, when x = h, by (1) ; 

also <j}(x) = 0, when x = 0. 

Hence by Eolle's Theorem, Art. 89, <j>'(x) == 0, for some value of x 
between and h. 

Calling this value of x, 6h, we have from (3) 

f'(a -f Bh) -R = 0. 

Substituting this value of R in (1), 

f(a + h)=f(a) + hf'(a + $h). 



EXPANSION OF FUNCTIONS 103 

92. Extension of Mean Value Theorem. We may extend the method 
of the preceding article so as to include the second derivative, and 

obtain f(a + h) =/(a) + hf (a) +|V" (* + Oh). 

Define i? by f(a + h)-f(a)-hf(a) ~M = 6 (1) 

Let $(x)=f(a + x)-f(a)-xf(a)-^R (2) 

Hence <f>'(x) =f (a + x) —f (a) — xE, 

4>"(x)=f"(a + x)-B (3) 

From (2) it is evident that <f>(x) = 0, when x = h, by (1); 
also <ft(x)=0, when a; = 0. 

Hence by Rolle's Theorem, Art. 89, <f>'(x) = 0, for some value, x 1} 
of x } between and h. 

Also <f>' (x) = 0, when x = 0. 

Hence <f>" (x) = 0, for some value, x 2 , between and x 1} that is, be- 
tween and h. Writing x 2 = Oh, we have from (3), 

f'(a + 0h)-E = O. 
Substituting this value of E in (1), we have 

f(a + h) =f(a) + hf (a) + ~f'{a + Oh). 

It is to be noticed that it is assumed that /(a;), f'(x), and /"(a;) are 
continuous from x = a to x — a + h. 

93. Taylor's Theorem. This may now be derived by extending the 
preceding method so as to include the ?*th derivative. 

It is assumed that fix) and its first n derivatives are continuous 
from x — a to x = a + h. 

Define It by 

f(a + h)-f(a)-kf(a)-gf»(a) ^/»-i( a )-f"i^0. (1) 

[2 \n-l \n 



104 DIFFERENTIAL CALCULUS 

Let 
+(x)=f(a+z)-f($-xfXa)-£f"(a) * /^( a )-^ 

if(x) = fr(a + x)-f'(a)-xf»(a) [ J^/^»( tt )- | -gL j» 



$«-* (») =/ w " 1 (a + a;) -/"- 1 (a) - a>2J, 

*-(oj)=/ l (a + aj)-iJ (2) 

As in the preceding articles, it is evident that 

cj)(x) = 0, when x = h, and also when x=0. 
Hence <f>'(x) = 0, when x = x 1 , where 0<x x <h. 
But <f>'(x) = 0, when # =0 ; hence 

<f>"(x) = 0, when x = x 2 , where < x 2 < x v 

Continuing this reasoning, we find 

<fr n (x) = 0, when x = x n , where < x n < x n _ u 
that is, where x n is between and h. 
Hence from (2) <£" (Oh) =f n (a + 6h) - R = 0. 

Substituting this value of R in (1), we have 
f(a + h)=f(a)+hf («)+! /"(«) + ... + |r^/-'(a)+^/"(a + fl»). 

Since a is ara/ quantity, we may write x in place of a, giving 

/(■ + A) =/(■) + V(«) +|/» + - + | J3 I /- 1 o*) 

+ .-/•(* + »). ..... (3) 



EXPANSION OF FUNCTIONS 



105 



94. Remainder. The last term of this equation 

I* 

is called the remainder after n terms in Taylor's Theorem. When the 
limit of this remainder is zero, as n is indefinitely increased, Taylor's 
Theorem gives a convergent series. 

We have already seen (Art. 86) that Maclaurin's Theorem is a 
special case of Taylor's Theorem, so that corresponding to (3) of the 
preceding article, we may write Maclaurin's Theorem 

/(*) = /(0) +.r/'(0) + |/"(0) + .- + -^_/..-'(0) +|/»(te), 

f(x) and its first n derivatives being assumed continuous for values 
from to sb. 

Thus the remainder after n terms in Maclaurin's Theorem is 

™» 



When 



Lim n=I=e 






= 0, 



(1) 



Maclaurin's Theorem gives a convergent series. 
Applying (1) to f(x) = e x , we have 



Lim, 



= 0, 



which is evidently satisfied for all values of x. 

The same is true for f(x) = sin x, and f(x) = cos x. 

llf(x) = log (1 4- *), (1) becomes 

Lhn_ \* (-1)-|!L=11 

U± (i + Bxy J 

=--[ ! =Ffe)-]=»- 

This is satisfied when |aj| < 1. 

It is to be noticed that the preceding test for convergence is of no 
practical use, unless the nth. derivative of f(x) can be expressed. 



CHAPTER IX 
INDETERMINATE FORMS 

95. Value of Fraction as Limit. The value of the fraction ^; ' for 

• , i ■■■■"# • <K«) ^> 

any assigned value ot x, as x =a, is ^ v y • 

This is a definite quantity, unless <j>(a) or \{/(a) is zero or infinity. 
When this is the case, we may, by regarding the fraction as a 
continuous variable, define its value when x equals a, as its limit 
when x approaches a. 

That is, the value of ^' ' , when x = a, is defined to be 

Lim x _ a ^ , or what is the same thing, 

Lim ±(a+M. 

h -°t(a + h) 

There is no difficulty in determining this limit immediately, when 
the numerator only, or the denominator only, is zero or infinity ; or 
when one is zero and the other infinity. 

We will now consider the cases where, for some assigned value of 
x, the numerator and denominator are both zero or both infinity. 
The fraction is then said to be indeterminate. 

96. Evaluation of the Indeterminate Form - • Frequently a trans- 
formation of the given fraction will determine its value. 

Thus, — j- — - — = - , when x = l. 

x 2 — l 

But if we reduce the fraction to its lowest terms, we have 

z 2 +.t-2 T . x + 2 3 

LinL,_, = Lim^i — [ — = - • 

* _1 x*-l Vhl 2 

106 



INDETERMINATE FORMS 107 



r — 2 

Again, — — = -, when x = 2. 

a z - 1 - 1 



By rationalizing the denominator, 



T . a; - 2 T . (a; - 2) (V» -1 + 1) 

Vai-1-1 g—'2 



= Lim^Va-l + 1) = 2. 

As another illustration, 

008 2g = °, whenfl = ?. 
cos 0- sin 4 

cos 2 (9 . cos 2 6- sin 2 (9 

"But Liui p _zr 7 : — 7 — Lim fl _zr ;r~ — : — w 

e ~ 4 cos 6 — sin (9 e ~4 cos^-siu^ 

= Lillian (cos 6 + sin 0) = cos- -f- sin-= -y^. 
4 4 4 

The Differential Calculus furnishes the following general method: 

97. Form <• yen- fraction, taking the derivative of the given numerator 
for << new numerator, and of the given denominator for a new denomi- 
nator. T/>>- value of this new fraction, for the assigned value of the 
variable, is /la- limiting value of the given fraction. 

We will now show how this rule is derived. 

Suppose the fraction ^ ■' —-, when x = a-, that is, <f>(a) =0, and 

= o. *W ° 

- By Art. 95 the required value of the fraction is the limit of 

. as h approaches zero. 
- h) 

By the Mean Value Theorem, (2), Art. 90 

(ft (a +h) = <t> (a) + h4'(a + Oh), 
^{a^h) = ^{a)- r h^(a-r-e i h) t 

where 6 and X are proper fractions. 



108 DIFFERENTIAL CALCULUS 

But since cj>(a) — and if/(a) = 0, we have 

<t>(a 4 h) = h<f>'(a 4 Oh) = <fr f (a + flft) t 
^(a 4 /i) ^'(a 4 dji) i}/'(a 4 0^) * 

Hence Lim ^^±£ = £M 

which is the theorem expressed by the rule. 

If <j>'(a) = and ij/' (a) = 0, it follows likewise that 

that is, the process expressed by the rule must be repeated, and as 
often as may be necessary to obtain a result which is not indeter- 
minate. 

For example, let us find the limiting value of the fraction in 
Art. 96. 

$(x) x 2 -l 0' 

y ,> / = ^ = -, when 05 = 1. 

xp\x) 2x 2' 

q 

Thus the required limiting value is - • 

Z 

For another example, let us find the limiting value, when x = 0, 

e * _|_ e -x _ 2 



1 — COS £C 




i// (a?) 1 — cos a? 


= -, when a; =0. 


<£'(<e) _ e x — e~* _ 
ij/'(x) sin a; 0' 


when x = 0. 


xj/"(x) cos a? 


when x = 0. 


Thus the required limiting value 


is 2. 



INDETERMIXATE FORMS 109 



EXAMPLES 

Find the limiting values of the following fractions for the assigned 
values of the variable. 

1. *0 g ~ 1 ;r~ 2 , whenz = 2. Ans. n + \. 

x- — 2 X L 

% log(3*» + s-3) whenx = l. Ans. 7. 

\ogx 

3. a ~ , when x = 0. Ans. log 6 a. 



.r— tan _1 .T 



vers fl 
sin 2 ' 






sin^ + |)-l 
log sin 2 



when x = 0. -4?is. — 2. 



5 108(^-4 8 + 6) when* = 2. ^n S . -2. 

log cos (# — 2) 

6 w-log(x + l) > wh enx = 0. 4«& g. 

XT Jj 



when = 0. Ans. 0. 



8. sinm "- sin "", whenm = «. Ans. cos no. 

sin (?n — n)« 



when0 = ?« ^4??s. - 



10. 5^=± 6 , when a =6. ^w. ^+J2S- 6 . 

a b —b" 1— log 6 

„ log >g -log.6 when a = 6. Ans. 



a — b b log b 

^-2ar' + 2x-l 



110 DIFFERENTIAL CALCULUS 



13> faums-ntans wh en z = 0. Ans. 2. 

nsmx — sin7ix 

, A tan nx — n tan x ■ -, ^ A xsec 2 ^— tana; 

14. — ; ; , wnen?i = l. Ans. — 

n sin x — sin nx sin x — x cos x 

- c m sin x — sin mx • , A A m s — m 

15. , when x = 0. Ans. 

(x — 1) e x + (a; + 1) e~ x 4 

16. eto + e " - 



log sec 2 x — af 



98. Evaluation of the Indeterminate Form °°. The method is the 

00 

same as that given in Art. 97 for the form - . 
It has been shown in that article that 

Lmu = o T^-Z L_ 1 = T-±-l , (1) 

if/ (a + li) i//(a) 

if <£(«)= 0, and ^(a)=0. 

It may be shown that (1) is true also, if <f>(a) = oo and \p (a) = oo . 
For the proof of this the student is referred to more extensive 
treatises on the Differential Calculus. 

For example, find the limiting value of ° x , when x = 0. 

cot x 

&£)=M^ = » when z = 0. 

\f/(x) cot x oo 

1 

6'(a?) a; sin 2 a; , A 

7?H= — = = a' whenx-0. 

y(x) — cosec^a? a; 

d>"(x) 2 sin a; cos a; A -, A 

,.) ( = = - = 0, when a; = 0. 

\\>"(x) 1 1. 

Thus the required limiting value is 0. 

oo 

Note. — ^The form — can in most cases be avoided by transformation into — 

GO J 



INDETERMINATE FORMS 111 

99. Evaluation of the Indeterminate Forms • x , cc — cc . 

Transform the expression into a fraction, which will assume either 

the form - or — • 
x 

For example, find the value of 

(?r — 2 x) tan x, when x = ~+ 
This takes the form • oc . 

But (tt-2x) taiix= 7r ~ 2x =®, when« = J. 

v J cot a; 0' 2 

^ = ~ 2 , = 2j when *«* 
i//(.r) -cosec 2 a- 2 

Thus the required limiting value is 2. 

For another example, find the limiting value of 

1 1 

log X X — 1 

This takes the form x — x 



, when x = 1. 



-p,-,4. 1 1 b — 1— log a; -, ., 

-But = ° . = - when x = 1. 

log sc x — 1 (a; — 1) log x 

^ = __£_=4 when a- = 1. 
'« 1-1+fcg. ° 

1 

„ = = - . when x = l c 

*(*) 1 + 1 2 ' 

a? 2 a; 

Thus the required limiting value is-- 



112 DIFFERENTIAL CALCULUS 

100. Evaluation of the Exponential Indeterminate Forms, 0°, l 00 , oo°. 
Take the logarithm of the given expression, which will have the 

form - or — • The limiting value of this logarithm will determine the 

oo 

given function. 

For example, find the limiting value of x*, when x = 0. 
This takes the form 0°. 

Let y = x x ; 

then log y = x log x = — ^— = — , when x = 0. 

x -\ GO 

1 

iLLA — = — x = 0, when x = 0. 

ij/'(x) _1_ 

x 2 
Thus the limiting value of log y is 0. 
Hence the limiting value of y is e° = 1. 

For another example find the limiting value of 

i 
(1 -f- ax)* t when x = 0. 

This takes the form of l 00 . 



i 
Let y = (l + ax)x, 



log y = lp g (! + ax ) = % when * = 0. 
a; 



y v y = — ! = a, when x = 0. 

^' (a) 1 



f (a) 

The limiting value of log y being a, the limiting value of y is e a 



INDETERMINATE FORMS 113 



EXAMPLES 

Find the limiting values of the following expressions for the 
assigned values of the variable. 

1. log(lH-aj), when a; = 0. Ans. 

x or 2 

3 

2. 2 X tan — , when # = 00. Ans. 3. 

2 T 

3. x tan v — j: sec x, when a; = ^. Ans. - 1. 

4. '°g tan "*, 

log tan &# 

5 » * 



2x? 2x tan a; ' 
sec 3 x 



6. 

sec o x 



„ f x 2 + x \-\ 
V 2 J : 

8. (cosec0) tan 



9. (tan0) cos< > 

2 sec 2 (9 - l\ tan2 ' 



10 



/ 2 sec 2 6 - I V 



c"\i 



12. (log a>>--, 

i 

13. (a z + z)x, 



«* / cos ax 4- cos bx \ xZ 
\ 2 J 



when a; = 


= 0. 


when a; 


= 00 


when a; : 


7T 

~2" 


when a' : 


= 0. 


when x = 


= 0. 


when x ■■ 


7T 

~2* 


when x 


= 1. 


when 6 


7T 

~2* 


when 0: 


7i' 

"2" 


when 0: 


_ 7r 
"4" 


when x 


= 0. 


when a; 


= e. 


when x = 


= 0. 


when a; = 


= 0. 



^±ns. 


6* 


Ans. 


5 
3' 


Ans. 


3 


Ans. 


vz 


Ans. 


1. 


Ans. 


4 

e 3- 


Ans. 


abc. 


Ans. 


i 


Ans. 


ae. 


Ans. 


1 



CHAPTER X 

MAXIMA AND MINIMA OF FUNCTIONS OF ONE 
INDEPENDENT VARIABLE 

101. Definition. A maximum value of a function is a value greater 
than those immediately preceding or immediately following. 

A minimum value of a function is a value less than those immedi- 
ately preceding or immediately following. 

If the function is represented by the curve y=f(x), then PM 
represents a maximum value of y or of /(»), and QN represents a 
minimum value. 



102. Conditions for a Maximum or a Minimum. 

It is evident that at both P and Q the tangent is parallel to OX, 
and therefore we have for both maxima and minima, 

clx 
Moreover, as we move along 
the curve from left to right, 
at P the slope changes from 
positive to negative ; but at Q, 
from negative to positive. 

In other words, 

At P the slope decreases as 
x increases. ... . (a) 

At Q the slope increases as 
x increases (b) 

By Art. 21 we have the case (a) 

d 




when 



dx 



(slope) < 0. 



• (1) 



114 



MAXIMA AND MINIMA OF FUNCTIONS L15 

But 



* (slope) = ^f^\ = % 
Iv * J dx\dxj lW 



dx 

and (1) becomes — \ < 0. 

efar 

Hence when ^ = 0, and ^ < 0, . (2) 

dx dx 2 

there is a maximum value of y. 

By similar reasoning we have the case (6), 

$y 
dx 2 



when — \ > 0. 



Hence when S = 0, and ^ > 0, (3) 

r/.v dar w 

there is a mi id mum value of ?/. 

For example, let us find the maximum and minimum values of 
the function « 

Put 2/== |_2o; 2 + 3^ + l. 

Then ^ = a*-4a?^& (4) 

da; 



Putting ( 4) = 0, s? - 4 a; + 3 = 0, 

whence 



fH*- 4 - • • • • < 5 > 



Substituting those values of a; in (5), we find 
when a =1, = _2<0; 



116 



DIFFERENTIAL CALCULUS 



Hence by (2) and (3), 
when x = 1, y has a maximum value ; 

when x = 3, y has a minimum value. 

From the given function we find 

that the maximum value of y is y = 2J, 

and the minimum value of y, y = 1. 



103. In exceptional cases it may happen that a value of x given by 

^ = 0, makes ^ == 0, so that neither (2) nor (3), Art. 102, is satisfied. 
dx dx- 

This would be the case for a 
point of inflection B (see Art. 
158) whose tangent is parallel 
to OX. Here the ordinate RL 
is neither a maximum nor a 
minimum. 

But there may be a maximum 
or minimum value of y, even 

when —4=0. This is more fully 
dx 2 J 

considered in Art. 106. The 
following article is also appli- 
cable to such cases. 




104. Second Method of determining Maxima and Minima. Maxima 

and minima may be determined from the first derivative — alone. 

d 2 dx 

without using — %. 

dx 2 

We have seen in Art. 102 that when y is a maximum, as at 

P, the slope, that is, — , changes from -f- to — ; and when y is a mini- 
ax 

mum, as at Q, -& changes from — to + . (It is understood that we 
dx 

pass along the curve from left to right.) 



MAXIMA AND MINIMA OF FUNCTIONS 117 

By examining the form of — , which should be expressed in factor 

form, we may determine whether it changes from -f to — , or from 
— to + , for any assigned value of x. 
Let us apply this method to the example in Art. 102, 

c ^ = X s - 4a? + 3 = - 1)(* - 3). 
dx 

Here — can change sign only when x = 1 or x = 3. 

dx ° ° J 

By supposing x to change from a value slightly less, to one slightly 
greater than 1, we find that (x — 1) changes from — to + ; but since 

the factor (x — 3) is then negative, it follows that — changes from 

dx 

+ to — , when x = 1, and denotes a maximum. In the same way, we 

find that — changes from — to +, when x = 3, and denotes a mini- 

dx 
mum. 

Again, consider the function y = (x — 4) 5 (# + 2) 4 . 

Differentiating and writing the result in factor form, 

c Il = 3 (3 x - 2)(x - 4:)\x + 2) 3 . 
dx 

"When x =-. -*- changes from — to + . 
3 dx 

"When x = — 2, -2 changes from + to — • 
dx 

When x = 4, -^ does woi change sign, 
dx 



since (a; — 4) 4 cannot be negative. 

2 
Hence we conclude that y is a minimum when # = -; a maximun 

o 

when x = — 2 ; but neither a maximum nor minimum when x = 4. 

As this method does not require —%, it is preferable to thai 

dx 2 

of Art. 102, when the second differentiation of y involves much work 



118 DIFFERENTIAL CALCULUS 

EXAMPLES 

1. Find the maximum value of 32 a; — x 4 . Ans. 48. 

Find the maximum and minimum values of the following functions. 

2. 2 X s — 3 x 2 — 12 x + 12. Ans. x = —l, gives a maximum 19. 

x = 2, gives a minimum — 8. 



3. 2 x 3 — 11 ic 2 + 12 x + 10. ^4ns. a? = -, gives a maximum 13-J-f- 



2 
x — 3, gives a minimum 1. 



4. « 3 + 9 (a — #) 3 . ^tfts. x = -— - , gives a maximum 



3 a _._ „ : 9 a 3 

4 ' 

3a . . ".-. 9 a 3 

# = — -, gives a minimum . 

4 16 



5. (x-l)(x-2)(x-3). 

A o 1 ■ • -2 

-4ws. a> = 2 = , gives a maximum 



V3 3V3" 

1 • ° 

x = 2 -\ = , gives a minimum 



V3 3V3 

6. 2 (3 a? + 2) 2 — 3 # 4 . -4ns. aj = 2, gives a maximum 80. 

7. Show that ^-^ — ^t — ^ ~ has no maximum nor minimum. 

x— 1 

8. ^_ + _^_ )W herea > 6. 

a a ~ x a 2 (a4-bY 

Ans. x = , gives a maximum ^ — — — '— 

a + b a 

a 2 . . (a-b) 2 

x = , gives a minimum ^ 1- 

a — b a 

1 no* / y* 

9. Show that the greatest value of — ^— is — . 

a; n ne 

9 

10. Show that the greatest value of cos 2 ■+- sin is -. 

8 



MAXIMA AND MINIMA OF FUNCTIONS 119 

11. Show that the maximum and minimum values of 

sin 2 $ + sin 2 f - — 6 ) are - and - . 



, 3 / 2 2 

12. Find the maximum value of a sin x + b cos x. Ans. Va 2 + b 2 . 

13. Find the maximum value of tan -1 x — tan -1 ~, the angles being 
taken in the first quadrant. . , _j3 

4' 

14. Show that the least value of a 2 tan 2 -f b 2 cot 2 is the same as 
that of ere** + &V", and equal to 2 ab. 

15. y = ^ a ~ x ) . u4?is. A minimum when x = - 
J a - 2 x 4 

16. 2/ = (.r-l)%r + 2) 3 . 

Ans. A maximum when a = ; a minimum when x = 1 ; 

neither when a; = — 2. 

17. 7/=0-2) 5 (2z + l) 4 . 

J./<s. A maximum when x = — - ; a minimum when x = — , 

2 lo 

neither when x = 2. 






105. Case where -^ = oo . It is to be noticed that -^ may change 
dx dx 

sign by passing through infinity instead of zero. 

Hence if ^=oo, 

dx 

for a finite value of # ; this value should be examined, as well as 
those given by 

''" =0. 
dx 



120 DIFFERENTIAL CALCULUS 

For example, suppose 

2 

y = a — b(x — c) 3 . 



Then 



hence we have 



dy_ 

dx 



25 



3(x-cy 
Y 



dy 
dx 



■■ oo , when x = c. 



It is evideDt that when x 



dx 
changes from 4- to — , indicating a 
maximum value of y, which is a. 

The figure shows the maximum 
ordinate PM, corresponding to a ' 

cusp at P. 

On the other hand, suppose y = a — b(x — c)" 3 ". 

b 




Then 



dy 
dx 



= oo when x = c. 



3(«-c) 8 



dy 



But as -^ does not change sign when x = c, there is no maxi- 
dx 
mum nor minimum. The corresponding curve is shown in the figure. 

Y 



MAXIMA AND MINIMA OF FUNCTIONS 121 

EXAMPLES 

Find the maximum and minimum values of the two following 
functions: 

1. y=(x + l)i{*-oy- 

Ans. A minimum when x = 5; a maximum when a; = ; 
a minimum when x = — 1. 



2. y = (2x-ay(x-a)*' 

Ans. A maximum when x = — ; a minimum when x = a. 



106. Conditions for Maxima and Minima by Taylor's Theorem. 

Suppose the function /(.r) to be a maximum when x = a. Then, by 
the definition in Art. 101, 

/(o)>/(a + ft), 

and also f(a) >f(a — h), 

where h is any small but finite quantity. Now, by the substitution 
of a for x in Taylor's Theorem, we have 

/(a + A)-/(a)= Rf(a) + ^/"(a) + i S /"'(a) + -. • (1)* 



[»' 



/(a-;0-/(a) = -7i/'(a) + f/"(a)-|V'» + -- • (2) 

By the hypothesis /(a + /*) — /(a) < 0, 

and also /(a — h)-f(a) < 0. 

Hence the second members of both (1) and (2) must be negative. 

* The rigorous form of Art. 93 may be used here without any change in the 
context. 



122 DIFFERENTIAL CALCULUS 

By taking h sufficiently small, the first term can be made numeri- 
cally greater than the sum of all the others, involving h 2 , h 3 , etc. 
Thus the sign of the entire second member will be that of the first 
term. As these have different signs in (1) and (2), the second mem- 
bers cannot both be negative unless 

/(a) = 0. 

Equations (1) and (2) then become 

f(a + A) -/(a.) = | /"(a) + |/"'(a) + ...» 

/(a-A)-/(a)=|/"(a)-|/''(a)+.... 

The term containing h 2 now determines the sign of the second 
members. That these may be negative, we must have 

f"(a)<6. 

If then /'(a) = and /"(a)<0, 

f(a) is a maximum. 

Similarly, it may be shown that if 

/'(a) = and /"(a)>0, 
f(a) will be a minimum. 

If /'(a) = and /"(a) = 0, 

similar reasoning will show that for a maximum we must also have 

f'"(a)=0 and f iv (a)<0; 

and for a minimum 

/"'(a)=0 and f lv (a)>0. 

The conditions may be generalized as follows : 
Suppose that 

f(a) = 0, /"(a) = 0, f"'(a) = 0, ... f(a) = 0, 

and that f n+1 (a) is not zero. 

Then if n is even, /(a) is neither a maximum nor a minimum. 
If n is odd, f(a) will be a maximum or a minimum, according as 
/ M+1 (a)<0 or >0. 



MAXIMA AND MINIMA OF FUNCTIONS 



123 



:>.i'(4-.t)(10-.t) 2 = 0, 



PROBLEMS IN MAXIMA AND MINIMA 

1. Divide 10 into two such parts that the product of the square 
of one and the cube of the other may be the greatest possible. 

Let x and 10 — x be the parts. Then x 2 (10 — x) 3 is to be a maxi- 
mum. Letting u = x 2 (10 — x) 3 , we find 

du 
dx 

from which we find that u is a maximum when x = 4. Hence the 
required parts are 4 and 6. 

2. A square piece of pasteboard whose side is a has a small 
square cut out at each corner. Find the side of this square that the 
remainder may form a box of maximum contents. 

Let x = the side of the small square. Then the contents of the 
box will be (a — 2 x) 2 x. Eepresenting this by u, we find that u is a 

maximum when x= -, which is the required answer. 

3. Find the greatest right cylinder that can be inscribed in a 
given right cone. 

Let AD = a, DC=b. 

Let x = DQ, the radius of the base of the cylinder, and y = PQ, 
its altitude. 

From the similar triangles ADC, PQC, 
we find 

b-x b y 



y=i« 



.r). 




The volume of the cylinder is 

7r:r?/ = 7r-X 2 (6 — x). 
a 



This will be a maximum when u^bxP—x 3 
is a maximum. 

This is found to be when x = | b, the radius of the base of the 
required cylinder. 

From this, y = -, the altitude of the cylinder. 
o 



124 



DIFFERENTIAL CALCULUS 



4. Determine the right cylinder of the greatest convex surface 
that can be inscribed in a given sphere. 

Let r — OP, the radius of the sphere ; x = OR, the radius of base 
of cylinder ; and y = PR, one half its altitude. 

From the right triangle OPR we have 

x 2 i -j g y 2 =r 2 . 
The convex surface of the cylinder is 



2 ttx • 2 y = 4 7rx Vr 2 — x 2 . 

We may put u equal to this expression, 
and determine the value of x that gives a 
maximum value of u. But the work may 
be shortened by the following considerations : 




4 irx Vr 2 — x 2 is a maximum, 



when 

and 

when its square 



x V?* 2 — # 2 is a maximum ; 
x-y/r 2 — x 2 is a maximum, 
i^x 2 — x 4 is a maximum.* 



Hence we may put u = r^x 2 — x k , from which we find u is a 



maximum, when x = 



V2 



From this y = — -, giving for the altitude of the cylinder, 

V2 

2?/ = rV2. 

Another Method. The equations 

The convex surface = 4 irxy, u = xy, (1) 

x 2 + y 2 = i*, (2) 

may be used without substituting in (1) the value of y from (2). 



* Since we are only concerned with the positive root of Vr 2 — x 2 . 



MAXIMA AND MINIMA OF FUNCTIONS 125 

Differentiating (1), ^L = y + X % (3) 

dx dx 

and differentiating (2). x + y c ^=0, * = _ 5 . 

dx dx y 

Substituting in (3), — = y = * 

dx y y 

Since x and y are positive quantities, it is evident that when 

x=y, — changes from -f to — , giving a maximum value of u. 
dx 

Combining x = y, with* (2), we have 

x = — = , y = — - , as before. 

V2 V2 

In some problems this method has some advantages over the first. 

5. Divide 48 into two parts, such that the sum of the square cd 
one and the cube of the other may be a minimum. Ans. 42f, 5-|. 

6. Divide 20 into two parts, such that the sum of four times the 
reciprocal of one and nine times the reciprocal of the other may be 
a minimum. Ans. 8, 12. 

7. A rectangular sheet of tin 15 inches long and 8 inches wide 
has a square cut out at each corner. Find the side of this square 
so that the remainder may form a box of maximum contents. 

Ans. 1J in. 

8. How far from the wall of a house must a man, whose eye is 
5 feet from the ground, stand, so that a window 5 feet high, whose 
sill is 9 feet from the ground, may subtend the greatest angle ? 

Ans. 6 ft. 

9. A wall 27 feet high is 8 feet from the side of a house. What 
is the length of the shortest ladder from the ground over the wall 
to the house ? Ans. 13 V 13= 46.87 ft. 



126 DIFFERENTIAL CALCULUS 

10. A person being in a boat 5 miles from the nearest point of the 
beach, wishes to reach in the shortest time a place 5 miles from that 
point along the shore; supposing he can run 6 miles an hour, but. 
row only at the rate of 4 miles an hour, required the place he must 
land. Ans. 929.1 yards from the place to be reached. 

11. Find the maximum rectangle that can be inscribed in an 
ellipse whose semiaxes are a and b. 

Ans. The sides are aV2 and &V2; the area, 2ab. 

12. A rectangular box, open at the top, with a square base, is to 
be constructed to contain 500 cubic inches. What must be its 
dimensions to require the least mate'rial ? 

Ans. Altitude, 5 in ; side of base, 10 in. 

13. A cylindrical tin tomato can is to be made which shall have 
a given capacity. Find what should be the ratio of the height to 
the diameter of the base that the smallest amount of tin shall be 
required. Ans. Height = diameter. 

14. What are the most economical proportions for an open cylin- 
drical water tank, if the cost of the sides per square foot is two 
thirds the cost of the bottom per square foot ? 

Ans. Height = f diameter. 

15. (a) Find the altitude of the rectangle of greatest area that 
can be inscribed in a circle whose radius is r. 

Ans. rV2; a square. 

(b) Find the altitude of the right cylinder of greatest volume that 

can be inscribed in a sphere whose radius is r. Ans. — ^« 

V3 

16. (a) Find the altitude of the isosceles triangle of greatest area 
inscribed in a circle of radius r. Ans. — ; equilateral triangle. 

(b) Find the altitude of the right cone of greatest volume inscribed 
in a sphere of radius r, Ans. — • 



MAXIMA AND MINIMA OF FUNCTIONS 127 

17. (a) Find the altitude of the isosceles triangle of least area 
circumscribed about a circle of radius r. 

Arts. 3r; equilateral triangle. 
(b) Find the altitude of the right cone of least volume circum- 
scribed about a sphere of radius r. Ans. 4 r. 

18. A right cone of maximum volume is inscribed in a given right 
cone, the vertex of one being at the center of the base of the other. 
Show that the altitude of the inscribed cone is one third the altitude 
of the other. 

19. Find the point of the line, 2x + y = 16, such that the sum 
of the squares of its distances from (4, 5) and (6, — 3) may be a 
minimum. Ans. (7, 2). 

20. Find the perpendicular distance from the origin to the line 
- -f i - = 1, bv rinding the minimum distance. Ans. 



a b ' J & ' ' Va 2 + &* 

21. A vessel is sailing due north 10 miles per hour. Another 
vessel 190 miles north of the first is sailing 15 miles per hour on a 
course East 30° South. When will they be nearest together, and 
what is their least distance apart ? 

Ans. In 7 hrs. Distance 15 Vo7 = 113.25 mi. 

22. A vessel is anchored 3 miles off the shore. Opposite a point 
5 miles farther along the shore, another vessel is anchored 9 miles 
from the shore. A boat from the first vessel is to land a passenger 
on the shore and then proceed to the other vessel. What is the 
shortest course of the boat ? Ans. 13 mi. 

23. The velocity of waves of length X in deep water is propor- 

7 

tional to \ - + -, where a is a certain linear magnitude. Show that 
* a X 

the velocity is a minimum when A = a. 

24. Assuming that the current in a voltaic cell is C = , E 

v -j- R 

being the electromotive force, r the internal, and R the external, 
resistance; and that the power given out is P=RC 2 \ show that P 
is a maximum when R = r. 



128 DIFFERENTIAL CALCULUS 

25. From a given circular sheet of metal, to cut out a sector, so 
that the remainder may form a conical vessel of maximum capacity. 

Arts. Angle of sector = ( 1 _ ^-)2 ir = 66° 14'. 



3, 

26. Find the height of a light on a wall so as best to illuminate 
a point on the floor a feet from the wall ; assuming that the illumi- 
nation is inversely as the square of the distance from the light, and 
directly as the sine of the inclination of the rays to the floor. 

Ans. 

V2 

27. At what point on the line joining the centres of two spheres 
must a light be placed, to illuminate the largest amount of spherical 
surface ? 

Ans. The centres being A, B ; the radii, a, b ; and P the required 
point; AP 2 :PB 2 =a?:b s . 

28. (a) The strength of a rectangular beam varies as the breadth 
and the square of the depth. Find the dimensions of the strongest 
beam that can be cut from a cylindrical log whose diameter is 2 a. 

Ans. Breadth = — . Depth = 2a A /?. 

V3 \3 

(6) The stiffness of a rectangular beam varies as the breadth and 

the cube of the depth. Find the dimensions of the stiffest beam 

that can be cut from the log. Ans. Breadth = a. Depth = a VS. 

29. The work of propelling a steamer through the water varies 
as the cube of her speed. Find the most economical speed against 
a current running 4 miles per hour. A?is. 6 mi. per hr. 

30. The cost of fuel consumed in propelling a steamer through 
the water varies as the cube of her speed, and is $ 25 per hour when 
the speed is 10 miles per hour. The other expenses are $ 100 per 
hour. Find the most economical speed. 

Ans. -^2000 = 12.6 mi. per hr. 

31. A weight of 1000 lbs. hanging 2 feet from one end of a lever 
is to be raised by an upward force at the other end. Supposing the 
lever to weigh 10 lbs. per fpot, find its length that the force may be 
a minimum. Ans. 20 ft. 



MAXIMA AND MINIMA OF FUNCTIONS 129 

32. (a) The lower corner of a leaf, whose width is a, is folded 
over so as just to reach the inner edge of the page. Find the width 
of the part folded over, when the length of the crease is a minimum. 

Ans. J a. 

(b) In the preceding example, find when the area of the triangle 
folded over is a minimum. Ans. When the width folded is -| a. 

33. A steel girder 25 feet long is moved on rollers along a pas- 
sageway 12.8 feet wide, and into a corridor at right angles to the 
passageway. ISTeglecting the horizontal width of the girder, how 
wide must the corridor be, in order that the girder may go around 
the corner ? Ans. 5.4 ft. 

34. Find the altitude of the least isosceles triangle that can be 
circumscribed about an ellipse whose semiaxes are a and b, the base 
of the triangle being parallel to the major axis. Ans. 3 b. 

35. A tangent is drawn to the ellipse whose semiaxes are a and 
b, such that the part intercepted by the axes is a minimum. Show 
that its length is a -f- b. 



CHAPTER XI 



PARTIAL DIFFERENTIATION 



107. Functions of Two or More Independent Variables. In the pre- 
ceding chapters differentiation has been applied only to functions of 
one independent variable. We shall now consider functions of more 
than one variable. 

Let u=f(x,y) 

be a function of the two independent variables x and y. 

Since x and y are independent of each other, we may suppose x to 
vary while y remains constant, or y to vary while x remains con- 
stant ; or we may suppose x and y to vary simultaneously. We 
must distinguish between the changes in u resulting from these dif- 
ferent suppositions. 

Let A x u denote the increment in u resulting from a change in x 
only, and A y u the increment in u from a change in y only. 

Let Aw, called the total 
increment of u, be the in- r N r ' 

crement when x and y both 
change. 

Suppose u the area of a 
rectangle whose sides are x 
and y. 

Then u = xy. 

If x changes from OA to u 
OA', while y remains con- 
stant, u is increased by the rectangle AM. 



Ay 


& 3 u 










C 




y 


u 




A^u 




x 




Ax 



That is, 



A x u = area AM. 
130 



obtain — • 



PARTIAL DIFFERENTIATION 131 

If y changes from OB to OB', while x remains constant, u is 
increased by the rectangle BX. 

That is, \u = area BN. 

If x and y both change together, we have for the total increment 
of u, A« = area AM+ area 5A"+ area MN* 

108. Partial Differentiation. This supposes only one of the inde- 
pendent variables to vary at the same time, so that the differentia- 
tion is performed by the same rules that have been applied to 
functions of a single variable. 

If we differentiate u = f(x, y), supposing x to vary, y remaining 

constant, we obtain — • 
dx 

If we differentiate, supposing y to vary, x remaining constant, we 

du 

dy 

The derivatives, — , — , thus derived, are called partial deriva- 
dx dy 

tives, and a special notatioD, — , — , is used for them. 

ox dy 

For example, if u = x 3 + 2 x?y — y 3 , 

— = 3 x 2 -f 4 xy, the ^derivative of u. 
dx 

— = 2x 2 — 3y 2 , the ^/-derivative of u. 

In general, whatever the number of independent variables, the 
partial derivatives are obtained by supposing only one to vary at a 
time. 

EXAMPLES 

Derive by partial differentiation the following results : 

du . du 

(to +„)*_(«.+«„)* 



1. 


V. 


xy 

•'• + y 




2. 


2 


= (ax 2 + 2 


bxy + off, 



132 DIFFERENTIAL CALCULUS 

o / \/ \/ \ du . du . du ^ 



- . dr dr 

4. r = A / T 2 , ?/ 2 w — = x — 

Vx +#> y dx dy 



5. m = log (x 3 + ace 2 ?/ + foci/ 2 + c?/ 3 ), a \- y — = 3- 

dx dy 



« ii , z . x du , du , du A 

6. % = £+- + -, a- +2/ +2 = 0. 

z x y ox oy dz 



vt e xy du . du , . iN 

7 - t *= x , , ' ^ + ^ = ( a; + 2/- 1 ) w - 

e x + e y ofl? a?/ 



8 . M = l 0g £^l + 2tan-^, *£ + *¥+ ^ = °- 

a; 4- 2/ 2/ ox oy x~ + y l 

10 . .. 1 J, gY + gl]f + glY, 1 

(^ + 2/ 2 + 2 2)i w vw V^y (^ + r+^-) 2 

11. 2 =log y aj4-log x y, a5logaj-^+ylogy-^ = 0. 

da; d?/ 



12. u= e z sin y + e y sin a?, 



^) 2+ (fr)* = e2x + e2 " + 2 e * + * sin (*+*)■ 



13. u = log (tan x + tan ^ + tan z), 



' o du , ■ o du . . o du 

sm 2x — + sin 2w f- sm 2z — =2. 

da; a?/ dz 



PARTIAL DIFFERENTIATION 



133 



109. Geometrical Illustration of Partial Derivatives. Let z = f(x, y) 
be the equation of the surface 
APCIL 

The ordinate PN is thus 
given for every point A" in the 
plane XT. 

Let APB and CPD be sec- 
tions of the surface by planes 
through P. parallel to XZ and 
YZ. respectively. 

If x and y both vary, P 
moves to some other position 
on the surface. y 

If x vary, y remaining con- 
stant, P moves on the curve of intersection APB. 

Hence — is the slope of the curve APB at P. 
dx 

If y vary, x remaining constant, P moves on the curve CPD. 
Hence — is the slope of the curve CPD at P. 




110. Equation of Tangent Plane. Angles with Coordinate Planes. 
In the figure of the preceding article, let P be the point (a?', y', z') ; 
PT, the tangent to APB in the plane APNM-, and PT, the tangent 
to CPD in the plane CPNL. 

It is evident from the preceding article that the equations of PT 
are 

, 6V 



dx 



-,(»-«'), y = y', 



and of PT, 



*"*'* #<*-&> X = X '- 



(1)* 
(2) 



. ^- denote the values of &-, $? , respectively, for (x', y' z'). 
ax' dy' dx dy 



134 DIFFERENTIAL CALCULUS 

The plane tangent to the surface at P contains the tangent lines 
PT and PT'. Its equation is 

z - z,==d 6 (x - x ' )+ W' {y ~ y,) (3) 

For (3) is of the first degree with respect to the current variables 
x, y, z, and is satisfied by (1), and also by (2). 

TJie equations of the normal through P are those of a line through 
(x', y', z') perpendicular to (3). Its equations are 

x — x'y — i/' , K ... 

- jn - = ?-^- = -(z-z') (4) 

d^_ dz^_ y ' v J 

dx' dy' 

The angles made by the tangent plane with the coordinate planes are 
equal to the inclinations of the normal to the coordinate axes. 

By analytic geometry of three dimensions, the direction cosines of 
the line perpendicular to (3) are proportional to 

dj_ dz[ _ 1 

dx' ' dy' ? 

Hence, if a, {3, y, are the inclinations of the normal to OX, OY, 
OZ, respectively, 

cos a _ cos /3 _ cos y 



dz; w 

dx dy' 



(5) 



Also cos 2 a + cos 2 /3 + cos 2 y = 1 (6) 



From (5) and (6) we find, dropping the accents, 

dz 

dx 



PARTIAL DIFFERENTIATION 135 






-(DM!) 

For the inclination of the tangent plane to XY, we have from (7), 

-"- i+ (D* + (IJ < 8 » 

The term slope used in geometry of two dimensions may thus be 
extended to three dimensions, as the tangent of the angle made by 
the tangent plane with the plane XY In this sense, 



=v(IJ + (SJ- 



EXAMPLES 

1. Find the equations of the tangent plane and normal, to the 

sphere x 2 — y 2 + z- = a 2 , at (V, y,' z'). 

dz _ x dz _ y t 
dx z' By z y 

„ dz' x' dz' y' 

Hence — = -, — ■ = — - ; - • • 

dx z' Oy z' 

Substituting in (3), z - z' = - "''- (x-x') - £ (y - y'), 

xx' + yy' + zz' — x' 2 + y' 2 -j- z' 2 = a 2 , vlws. 



136 DIFFERENTIAL CALCULUS 

From (4) we find for the normal 



^-^-,= (y-y')- = ^-^ 

x y 

— — 1 = ^ — 1 = - — 1, - = 2- = -. Ans. 

x' y' z' x y' z' 

2. Find the equations of tangent plane and normal to the cone, 

3x 2 -y 2 + 2z 2 = 0, at (x', y\ z'). 

Ans. 3xx'-yy'+2zz' = 0, ^-^ = y -^- = z ~ z ' ■ 

3 a;' —y' 2z' 

3. Find the equation of the tangent plane to the elliptic parabo- 
loid, z= 3 x 2 + 2 y 2 , at the point (1, 2, 11). 

Ans. 6x+8y — 2 = 11. 

4. Find the equations of tangent plane and normal to the ellipsoid, 

x 2 + 2y 2 + 3z 2 =20, 
at the point x = 3, y = 2, z being positive. 

Ans. 3x + ±y + 3z = 20; x = z + 2, 3y = 4;Z + 2. 
Find the slope of this tangent plane. Ans. -§. 

5. Find the equation of the tangent plane to the sphere, 

x 2 + y 2 + z 2 -2x + 2y = l, at (x', y', z'). 

Ans. icaj'-f- yy' + zz' — x — x' + y + y' =zl. 

111. Partial Derivatives of Higher Orders. By successive differ- 
entiation, the independent variables varying only one at a time, we 

may obtain 

d 2 u d 2 u dhc dhc 

dx 2 ' dy 2 ' dx* dy 4> 

If we differentiate u with respect to x, then this result with respect 

to y, we obtain — I -^ \ which is written — - — -. 
dy\dxj dydx 



PARTIAL DIFFERENTIATION 137 

Similarly, — is the result of three successive differentiations, 

J dyd.r 

two with respect to x, and one with respect to y. It will now be shown 
that this result is independent of the order of these differentiations. 

*\ n 

In other words, the operations — and — are commutative. 

dx dy 

. That is, 



dydx dxdy dydx 2 dxdydx dx*dy 

112. Given u=f(x,y), (1) 

to prove that «fe)"Sfe> 



Supposing x to change in (1), y being constant, 

Au = f(x + As, y) -f(x, y) ^ 

Ax " Ax 



(2) 



Now supposing y to change in (2), x being constant, 

AVAtA = f( x + As, y + Ay) -f(x, y + Ay) -/(a? + As, y) +f(x, gfl 
Ay\Ax) Ay Ax 

Reversing the above order, we find 

A» = f(x, y + Ay) -f(x, y) and 
Ay Ay 

A (*^\ = f( x + Ax > y + A y) -f( x + Aa: > y) — f( x > y + A ^) + f fa y) . 

Ax\Ay) Ax Ay 

Hence ±/A«\ A/Aii\ (3) 

Ay\AxJ Ax\Ay) J 

The mean value theorem, (2), Art. 90, may be expressed in the 

form — = f'(x+0>Ax), where u =f(x). 0< 6< 1. 
As 



138 DIFFERENTIAL CALCULUS 

In the present case, where u =f(x, y) f 

ty{lx) = iy fx {X + &1 ' AX ' V) =fyx(X + ^' AX ' V + $rAy) ' 

Similarly, ^=f( x ,y + B^Ay), 

Ay 

and Ax\^r) = f* y<KX + 6i ' Ax> V + ° 3 '^' 

By (3) /^(aj+^.Ax, y + O r Ay) =f xy (x + 6fAx, y + 8 .Ay). 

Taking the limits as Aaj, A?/, approach zero, and assuming the 
functions involved to be continuous, 

fyx(%,y)=fxy(%,y)- 

mi , • d fdu\ d fdu\ d 2 u d 2 u 

That is, - — =-— or _—_ = _-— . 

ay\axj dx\&yj ay ax ax ay 



This principle, that the order of differentiation is immaterial, may 
be extended to any number of differentiations. 

Thus d 3 u = d 2 fdu\ = d 2 fdu\ = &u 

dydx 2 dydx\dx) dxdy\dxj dxdydx 



= d f d 2 u \ d f 



d 2 u \ d 3 u 



dx\dydxj dx\dxdyj dx 2 dy 

It is evident that the same is true of functions of three or more 
variables. 

dx ay 



PARTIAL DIFFERENTIATION 139 



EXAMPLES 

A erlfy = — — in Lxs. l-o. 

dx dy dy dx 

1. u = ax + l) ! / , 2. w = fBylog?. 3. ?* = (a; + y) e*-». 

ay + bx y ■ 

Derive the following results : 

d 4 a d 4 u d*u 



4. u = a.x* -f 6 Saf^ 8 + c?/ 4 , 



dx 2 dy 2 dx dy dx dy dx dy 2 dx 



s. .=iog(*» +! o, g+g=°- 

6. ,=(3. + ,)3 + sin(2,-,), find g_^_«g. 

- a* 2 , , x n ■> *d 2 u , c, dhi . odhi 

7. tt = - + log-, find ar— -f 2a^-— - + 2/ 2 — 

1/ y dar oxdy dy z 

8. , ■ = ,= tan-' 2 - r tan- 5, -^ = ££ 

n / ,. , „\ a d 2 q , 1 dq . 1 d 2 q A 

d> 2 r 5?' r 2 d#- 

10. u = log ( e * + e" + <*) t4V = 2e^+- 3 «. 

dxdydz 

11. i^ztan 1 -, + + = o. 

?/ dar dy dz- 

10 i /".■>, ^ d 2 ii . d 2 u , d 2 u 1 

12. u = log (ar -f y 2 + z 2 ), — : + t-= + 



dx 2 dy 2 dz 2 x 2 -\-y 2 -\-z 2 

* " ? 1%, * V * 

io 999,00'). 999 U IV 0,9,0 

13. u - y-z-e- + z-.cre- + afyV, ^ ^ = e- + er + e 2 . 

14. u = sin (jj + z) sin (z + a;) sin (x -f y), 

2cos(2a; + 2?/ + 2z). 



dxdydz 



140 DIFFERENTIAL CALCULUS 

113. Total Derivative. Total Differential. In Art. 107 we have 
referred to the change in u when x and y vary simultaneously. 
This change is called the total increment of u. Thus the total incre- 
mentof u=f(x,y) 

is Au = f(x + Ax, y + Ay)~ f{x, y). 

The terms total derivative and total differential are also used. For 
example, 
let u = x s y-3x 2 y 2 , (1) 

and suppose x and y to be functions of a variable t. 
Differentiating with respect to t, 

dt dt K y) dt K J) 

= tfty. + 3^2 dx _ 6x2 dy _ 6 2 dx 
dt * dt J dt y dt 

= (Zx 2 y-$xtf)f t +(x*-(Sx 2 y) d X .... (2) 



But from (1) we find 



Cj7/ fi?f 

— = 3 x 2 y — 6 xy 2 , — = x z — 6x 2 y. 
ox dy 

So that (2) may be written 

du _du dx du dy ,q\ 

dt~dxdtdydt' r ^ ' 

If we had used differentials in differentiating (1) we should have 
obtained 

du = — dx H dy (4) 

dx dy J w 

dn 

— in (2) and (3) is called the total derivative, and du in (4) the total 

aii 

differential, of u. 

We proceed to show that (3) and (4) are true for any function of 
x and y. 



PARTIAL DIFFERENTIATION 141 

Noticing that An is the total increment of u, 

and A x u, A L/ u, the partial increments, when x and y vary separately, 

let 

u = f(x, y), x and y being functions of t. 





u'=r\x + Ax,y), 




u"=f(x + Ax,y + Ay). 


Then 


A x m = u' — it, 




A y u' = w" — m', 




Ait = u" — m. 


Hence 


Am = A x m + A,.M r , 


and 


Am _ A x u A.U A„w' Am 
At ~~ Aa A£ Ay At " 



Taking the limits of each member, as At, and consequently Ax, Ay, 
approach zero, clu = dudx du dy (5) 

dt dxdt dy dt' K } 

since the limit of u' is u. 

This may be written in the differential form 

du = — dx + — dy (6) 

dx dy J w 

In the same way, if u =f(x, y, z), where x, y, z are functions of t, 

dt ~ dx dt dy dt dz dt' ^ 

and ■ ri„ = ^rl< + ^dy + ^dz (8) 

ox dy dz 

We may write in (8) 

du , 7 cht , , du ■, 1 

— tfaj = d, tt _ dy = d„w, — - cZz = d z u, 

ox dy dz 

giving du = d x u + d v u 4- d z u, 

that is, the total differential of m is the sum of its partial differentials. 



142 DIFFERENTIAL CALCULUS 

This principle, as expressed by du = d x u + d y u, may be illustrated 
by the figure of Art. 107, from which we have 

Aw = A x u + A y u + area MN, 

that is, Aw == A x w + A u u -+- Ax Ay. 

As Ax and Ay approach zero, the last term diminishes more rap- 
idly than the others, and we may write 

Aw = A x u + A y u, K approximately, 

the closeness of the approximation increasing as Ax and Ay 
approach zero. 

If in (5) we suppose t == x, 
then u=f(x, y), y being a function of x ; 

and (5) becomes du^du + 3udy_ (9) 

dx ox By dx 

Similarly, if in (7), t =.x } 

u =f(x, y, z), y and z being functions of x ; 

, du du . du dv , du dz /iAN 

whence — = — + -h -= (10) 

dx dx dy dx dz dx ' 

EXAMPLES 

Find the total derivative of w by (5) or (7) in the three following : 

1. u=f(x,y,z), where x = t 2 } y = 1?, z—-. 

— — 2 t — 4- ^ t 2 — —- — 
dt dx dy t 2 dz' 

2. w = log (x 2 ~ y 2 ), where x = a cos t, y = a sin t. 

^ = - 2 tan 2 t. 
dt 

3- w = tan -1 -, where x = 2t, y = 1 — t 2 . — = — 

y r .r dt l + t\ 



PARTIAL DIFFERENTIATION 143 

Apply (10) to the two following : 
4. u =/(*, y, z), where y = x 2 — x, z = x? — x 2 . 

( h = ^+(2x-l)^ + (3x*-2x)^. 
dx dx v dy K ) dz 

o . u = tan -» , where y = d— ar, 2 = 1 — o flr, 



da? 1 + x 2 ' 

Find the total differential by (6) or (8) in the following : 
6. u = ax 2 + 2 6«y + cy 2 , du = 2 (ax + &?/) da; + 2 (6aj + c#) dy. 

v » y 

8 u = l r r sin i( x + y) j du = sinydx-smxdy m 

sin i (a? — y)' cos a) — cos ?/ 

9 . v. = ax 2 + ty 2 + cz 2 + 2/#2 + 2 gr«aj + 2 fta#, 

du =2 (ax + hy+gz) dx + 2(hx + by+fz) dy + 2(gx +Jy + cz)dz. 

10. it = af' du = a?* -1 (yz cZx + zx log a; c?y + xy log a; dz) . 

11. ?/ = tan 2 a:tair?/tairz, du = 4uf-^— + -^_ + __^_\ 

\sin2a; sin2y sin 2zy 

If the variable t in (5) and (7) denotes the time, we have the re- 
lation between the rates of increase of the variables. 

For illustration consider the following example : 

12. One side of a plane triangle is 8 feet long, and increasing 4 
inches per second ; another side is 5 feet, and decreasing 2 inches 

cond. The included angle is 00°, and increasing 2° per second. 
At what rate is the area of the triangle increasing? 



144 DIFFERENTIAL CALCULUS 

The area A = -be sin A, from which 

dA c • ,db , b • A dc . be A dA 

— = - sin^l V- - sm^l cos A — 

dt 2 dt 2 dt 2 dt 

= ^sin^.i + 5sin^.-i + ^cos^-^ 
2 32 6 2 90 

= .4934 sq. ft. = 70.05 sq. in. per sec. 

13. One side of a rectangle is 10 inches long, and increasing uni- 
formly 2 inches per second. The other side is 15 inches long, and 
decreasing uniformly 1 inch per second. At what rate is the area 
increasing ? Ans. 20 sq. in. per sec. 

At what rate after the lapse of 2 seconds ? 

Ans. 12 sq. in. per sec. 

14. The altitude of a circular cone is 100 inches, and decreasing 
10 inches per second, and the radius of the base is 50 inches and 
increasing 5 inches per second. At what rate is the volume in- 
creasing ? Ans. 15.15 cu. ft. per sec. 

15. In Ex. 12, at what rate is the side opposite the given angle 
increasing ? Ans. 8.63 in. per sec. 

114. Differentiation of an Implicit Function. (See Art. 66.) The 
derivative of an implicit function may be expressed in terms of 
partial derivatives. 

The equation connecting y and x, by transposing all the terms to 
one member, may be represented by 

+ (*,y)=o (i) 

Let u = cj>(x, y). 



From (9), Art. 113, we have for the total derivative of u, 

du _ du da dy 
dx dx dy dx 



PARTIAL DIFFERENTIATION 145 

But by (1) x and y must have such values that u may be zero, that 

is, a constant ; and therefore its total derivative— must be zero. 

dx 



Hence 



and ^ = _-_ (2) 



du 
dx 


du dy _ 

By dx 


= 0, 




dy = 


du 
_ dx 

du 






dx 








dy 






•om 




x?y 2 + ^y 3 = 


--a 5 . 



For example, find — from 
dx 



Let u = xPy- -+- xry 3 — a 5 . 



J? - 3 a^ + 2 a-r, f? = 2 &y + 3 *y. 

o.r dy 



dy = 3.i-y+ 2ay» = 3ay + 2y 2 
cfa 2^ + 3a^?/ 2 " 2z 2 + 3a*/' 



By (2) 

In the same way find the first derivatives in the examples of Art. 67 



115. Extension of Taylor's Theorem to Functions of Two Inde- 
pendent Variables. If we apply Taylor's Theorem 

to f(x+7i,y+k), 



regarding x as the only variable, we have 
f(x+ h, y + k) =f(x, y+k) + h~f(x, y + k) 



146 DIFFERENTIAL CALCULUS 

Now expanding f(x, y + 7c), regarding y as the only variable, 

d k 2 d 2 

f(x, y + k) =f(x, y) + k — f(x, y) + - y 2 f(x, y) + .... 



Substituting this in (1), 

fix + h,y + k) =f(x, y)+h±f(x, y) + h^-f(x, y) 



+ i /i2 £* /( ^ +2J ^ f(x > *>+"£/<*« 



+ .». (2) 



This may be expressed in the symbolic form thus : 
f(x + h,y + Jc) =f(x, y) + fhf x + kjj-\ f(x, y) 



m, as 



where (h h& — ) is to be expanded by the Binomial Theore 

V dx dyj 

if h— and k~ were the two terms of the binomial, and the result- 
dx dy 

ing terms applied separately tof(x } y). 

116. Taylor's Theorem applied to Functions of Any Number of In- 
dependent Variables. By a method similar to that of the preceding 
article we shall find 

f(x + h,y + k,z^l)=f(x,y,z) + (lij- + kj- + l£^f(x,y y z) 

+ |(*s + *S +, s) i ^^ f > +: - 

This expansion may be extended to any number of variables. 



PARTIAL DIFFERENTIATION 147 

EXAM PLES 

1. Expand log (a? + h) log (y + A). 

T . -, N i i d>/ los: y d?< losr x 

Let u =/(.i\ y) = log x log y, — = — a^ , — = -£-, 

dx a; ay ?/ 

d 2 ?? _ log?/ 3 2 m _ 1 d' 2 u _ log re 
d.r ar ? dyda; scy* dy~ y 2 

By (2), Art. 115, log (»+*) log (y + A) = log a logy 

. L , A* i A 2 , . AA A 2 , 

4- -logy + -loga-- _ log?/ H 5 _ i loga5+— - 

a; 2/ *& xy 2y- 

2. (a + A) 3 (y + A) 2 = xY + 3 A.r 2 y 2 + 2 fc^y 

+ 3 h 2 xy 2 + 6 AAa; 2 y + A~V + • • •• 

3. sin [(a: + A) (y + A)] = sin (xy) + Ay cos (xy) -f A» cos (#?/) 

7 2 2 V, 9 ,2 ' 

- ^-sin (xy) + 7ik [cos (xy)-xy sin (a-y)] - — sin(a;y) + •••. 



CHAPTER XII 
CHANGE OF THE VARIABLES IN DERIVATIVES 



117. To express 



dy d 2 y d 3 y 



••in terms of 



dx d 2 x d 3 x 



dx dx* dx 3 ' dy dy 2 ' dy 3 ' 

This is changing the independent variable from x to y. 



By (1), Art. 56, 
By (3), Art. 56, 
From (1), 



Similarly, 



From (2), 



dy = ±_ 

dx dx 

dy 

dry_d L dy_d_dy dy 
dx 2 dxdx dydx dx 

d 2 x 
d dy _ d 1 dy 2 

dy dx ~ dy dx ~ (ax* 
dy [dy, 

d?x 

d 2 y _ dy 2 
dx 2 ~ fdx\ 3 

. W 

dPj[_d L d^_d L d^y dy 
dx 3 dx dx 2 dy dx 2 dx 

fd 2 x\ 2 _ dx (Px 
d d 2 y \dy 2 ) dydy 3 



dy dx 2 



dx^ 4 
dy 



fdrx^ 2 _ dx d 3 x 
d 3 y \dy 2 ) dydy 3 



dx 3 



dx^ 5 
148 



(1) 



(2) 



(3) 



CHANGE OF THE VARIABLES IN DERIVATIVES 149 

It is sometimes necessary in the derivatives, 

dy d*y effy 

dx dar dx? 

to introduce a new variable z in place of x or y, z being a given 
function of the variable removed. 

There are two cases, according as z replaces y or x. 

118 First. To express 4, g, g, -, in terms of * * * ••, 
dx dxr dx 6 dx dxr dxr 

where y is a given function of z. 

By (3), Art. 56, * = **. 

J v " ' dx dz dx 

d 2 y _ d fdy\dz dydh__ d 2 y/dz\ 2 , <fy d?z 
dx 2 dx\dz Jdx dz dx 2 dz\dxj dz dx 2 ' 



Similarly, we find 

d 3 y _ cPy/dzY, o^ydz_dh,dy d?z_ 
dx 3 dz 3 \dx) dz 2 dxdx 2 dz dx 3 ' 

Similarly, — ^, —4, •-., may be expressed in terms of z and x. 

(XX CIX 

It is to be noticed that in this case there is no change of the in- 
dependent variable, which remains x. 
For example, suppose y = z*. 

Then *V = 3z 2 ^. 

dx dx 

dx 2 \dx) dx 1 



^ = 6 f*Y+ 18**^ + 3*^. 
dx 3 \dxj dxdx 2 dx 3 



150 DIFFERENTIAL CALCULUS 

119. Second. To express^, **, *M , ..., 

dx 9 dx 2 ' dx 3 ' ' 

in terms of -2 — a _Jl • where a; is a given function of z. 

dz dz 2 dz 6 

This is changing the independent variable from x to z. 
By (3), Art. 56, 





dy 


dy _ dy dz _ 
dx dz dx 


dz 
dx 




dz 




d fdy\ 


d*y = d L fay 
dx 2 dz \dx y 


\dz _dz\dx) 
)dx~ dx 




dz 


dx d 2 y 


dy d 2 x 


dz dz 2 


dz dz 2 






Similarly, higher derivatives may be expressed. In practice it 
is generally easier to work out each case by itself. 
For example, suppose x = z 3 . 





dy _ dy dz 
dx dz dx 


But 


dx _ 2 dz _ z~ 2 
dz~ Z ' ~dx~~3' 


Hence 


ySdy_l z - 2 <ty 




dx 3 dz 



(1) 



d 2 y _ d_ fdy\ _ d_ fdy\ dz 
dx 2 dx\dx) dz\dxjdx 

From(l), *(&\u.±(f**!-i r *®\. 

v " dz\dx) S\ dz 2 dz) 

Hence '&*(,-««& _2r*#\ (2) 

dx 2 9\ dz 2 dz) w 



CHANGE OF THE VARIABLES IN DERIVATIVES 151 

Similarly, S-#(§V£. 
dar dz \darj dx 

From (2), */*3t\ = */V^- 6z^^+ 10z~^ 

v h dz\dx 2 J 9\ dz" dz 2 dz 

Hence SU A(r*S*- 6r^+10 *+$L\ 

dx 3 27 V c?2 3 dz 2 dz 1 



EXAMPLES 

Change the independent variable from x to y in the two following 
equations : 

1. 3 (W- *£* - *W= 0. 4»s. *5 + & = 0. 

yiry dsccfcc 3 dx\dxj dy s dy 2 

\ c?oj J\dx~J \ dx Jdx dx s 

Ans. (**\U** + a y* 

\dy-J \dy Jdif 

Change the variable from y to z in the two following equations : 

3 «fff-i , 2(i+ y y %Y „_ tan/ 
3 ^~ 1 + TTFW' y 

iK ^-2^Y=eos**. 
da; 2 Veto:/ 

v 'cfo 3 dxdx 2 

Change the independent variable from x to z in the following 
equations : 

'/.' ,J ./' 'A'' (for C?Z 

das' 1 + a^rfx (1 + z 2 ) 2 ciz J J 



152 DIFFERENTIAL CALCULUS 

7. (2aj-l) 3 g + (2a J -l)^ = 2 2 /, 2» = l + e'. 

^ s . 4^-12^-f-9^-2/ = 0. 
dr dr dz 

8. ^^ + 6^ + 9^ + 3^ + 2/ = ^^, a; = e*. 

da; 4 dx 3 dar da; 

^ d? + d** +2/ -*' 



120. Transformation of Partial Derivatives from Rectangular to Polar 
Coordinates. 

Given u — f(x, y), 

to express — - and — in terms of — and — , where x, y, are rec- 
dx dy dr dO 

tangular, and r, 6, polar coordinates. 

We have from (5), Art. 113, regarding u as a function of r and 0, 

du _ du dr , du dO ,^ . 

dx dr dx d6 dx 

du_dudr dudO ^x 

dy~drdy <j$dy U 

The values of — , — -, — , — , are now to be found from the rela- 
dx ay ox dy 

tions between a;, y, and r, 0. 

These are a; = rcos0, y = rsmO (3) 

But in the partial derivatives — ^, — ^ , and — , — , r and 6 are re- 
da; dy dx dy 

garded as functions of x and y. 

These are, from (3), 

^ = a^ + 2/ 2 , d = tan- 1 ^. 
x 



CHAXGE OF THE VARIABLES IN DERIVATIVES 153 

Differentiating, we find 

dx r 8y r 

86 _ y _ sin 6 86 _ x _ cosfl 

dx x 2 + y- r ' 8y x 2 + y 2 r 

Substituting in (1) and (2), we have 

8u » 8u sin 6 8u //IN 

— = cos0- — , (4) 

dx dr r 86 



8u • * 8u . cos 6 8u /trx 

— = sm0 — + — (5) 

8y dr r 86 



121. Transformation of — A from Rectangular to Polar Co- 

dxr 8y 2 

*\ 

ordinates. By substituting in (4), Art. 120) — for u, we have 

dx 

8rv L _ 8_(8u\_ q 8 (8u\_ sin 6 8 (8u\ ,*. 

dx*~dx\dxj C ° S 8i\dx) ~8~6\8x) ^' 



Differentiating (4), Art. 120, with respect to r, 

8 (8v\_ q8 2 h sinfl 8 2 u sin fl da ,o\ 

8r\8x)~ dr 2 r 8r86 r 2 8$ K) 



Differentiating (4), Art. 120, with respect to 6, 

86\8x) drd6 dr r 86 2 r 86 W 



154 DIFFERENTIAL CALCULUS 



Substituting (2) and (3) in (1), we have 

dhi 2 a d 2 u 2 sin cos dhi , sin 2 d 2 u , sin 2 du 

— = cos 6 1 1 

dx 2 dr 2 r drdd r 2 d$ 2 r dr 



d$ 



Adding (4) and (5^ =*>* obtain 

c% dhi _ 5 

dx 2 dy 2 dr 2 r dr r 2 d& 



^ 2 u . d 2 u d 2 u . 1 du . 1 d 2 u 






2 sin cos du ... 



Similarly by using (5), Art. 120, instead of (4), we find 

d 2 u _ • 2 a dhi 2 sin cos d 2 u cos 2 d 2 u cos 2 6 du 
dy 2 dr 2 r drdO r 2 dO 2 r dr 

2 sin cos du 



.(5) 



CHAPTER XIII 

MAXIMA AND MINIMA OF FUNCTIONS OF TWO OR MORE 
INDEPENDENT VARIABLES 

122. Definition. A function of two independent variables, f(x, y) } 
is said to have a maximum value when x = a, y = b; when, for all 
sufficiently small numerical values of h and k, 

f(ti,b)>f(a + h,b + k), (a) 

and a minimum value, when 

/(a, 5)</(a + *»& + &). (6) 

123. Conditions for Maxima or Minima. 
If u=f(x,y), 

we find that a necessary condition for both (a) and (6) is that 

— = 0, and — = 0, when x = a, y = b. 
ox dy 

This may be shown as follows : 

Conditions (a) and (&) must hold when Jc = 0, and we have for a 
maximum 

f(a,b)>f(a + h,b), 

and for a minimum 

f(a, 6)</(a + ft, &), 

for sufficiently small values of //. 

155 



156 DIFFERENTIAL CALCULUS 

We thus have for consideration a function of only one variable. 
By Art. 106, we must have for both maximum and minimum, 

— f(x, b) = 0, when x = a, 
dx 

that is jrf( x ' ^ = ®> wnen x = a > V =b> 

ox 

Similarly, by letting h — in (a) and (6), we may derive 
~\ 

T f(%, V) = 0, when a; = a, y = b. 
dy 

These conditions for a maximum or minimum are necessary but 
not sufficient. As in the case of maxima and minima of functions 
of one variable, there are additional conditions involving derivatives 
of higher orders. These we shall give without proof, as their 
rigorous derivation is beyond the scope of this book. 

The conditions for a maximum or minimum value of u = f(x, y) 
are as follows : 

For either a maximum or minimum, 

£ = 0, and |5_0s (1) 

ox dy 

also 6S-Y<SS < 2 > 

\dy ox J dx 2 dy 2 



S<* and w 



For a maximum, -^ < 0, and — ^ < (3) 



For a minimum, — - > 0, and — > (4) 

ox 2 dy 2 

124 Functions of Three Independent Variables. The conditions 
for a maximum or minimum value of u =f(x, y, z) are as follows : 
For either a maximum or minimum, 

£?!f — o ^ — ^ u — 
dx ' dy ' dz 



and (^L\< 

\dx dy) 



d 2 u d 2 u 
dx 2 dy 2 



MAXIMA AND MINIMA OF FUNCTIONS 



157 



For a maximum 



d 2 u 



dx 2 



-<0, and A<0; 



for a minimum, 



|^>0, and A>0; 



where A = 



d 2 u d 2 u dh 



dx 2 * dx dy dx dz 
d 2 u d 2 u d 2 u 



dx dy dy 2 ' dy dz 
d 2 u d 2 u d 2 u 



dx dz dy dz dz 1 



EXAMPLES 
1. Find the maximum value of 

u = 3 axy — xP — y 3 . 



Here ^ = 3ay-3x 2 , —=3ax-Sy 2 . 

dx dy 



» n d 2 U a d 2 U a d 2 U 

Also ^j =_6flj > -z-i = - 6 y> r~^- =Sa " 

dx 2 dy 2 dx dy 



Applying (1), Art. 123, we have 

ay — x 2 = 0, and ax — y 2 = 0; 



whence 



x = 0, y = 0; or x = a, y = a. 



The values x = 0, y = 0, give 



t-, = 0, — , = 0, — — = 3ct, 
dx- dy dx dy 



which do not satisfy (2), Art. 123. 

Hence they do not give a maximum or minimum. 



158 DIFFERENTIAL CALCULUS 



The values x — a, y = a, give 

d 2 u n d 2 u a d 2 u o 

- — -• = — 6 a, — - = — ba, = 6 a, 

dx 2 by 2 dx By 

which satisfy both (2) and (3), Art. 123. 

Hence they give a maximum value of u, which is a 3 . 



2. Find the maximum value of xyz, subject to the condition 

t + vl + t 1 (1) 

a 2 fr c- 

c 2 a 2 b 1 

and as xyz is numerically a maximum when x^yh 2 is a maximum, 
we put 

. A 2a.- 2 2« 2 \ 



50a52/ 

From — = and — = 0, we find, as the only values satisfying 
dx dy 



(2), Art. 123, 



x = — — , y = — — which give 

V3 V3 



^ 9 ' fy 2 9 ' da% 9 ' 



MAXIMA AND MINIMA OF FUNCTIONS 159 

As these values satisfy (2) and (3), Art. 123, it follows that xyz is 
a maximum when 

, _ a _ b c 

a 3 V3 ~ V3 

The maximum value of xyz is — 

3v3 

3. Find the values of x, y, z that render 

ar + y 2 + z 2 + x — 2 z — xy 

a minimum. A 2 1 i 

Ans. #=— -, y=—-, 2 = 1. 

3 3 

4. Find the maximum value of 



(a — x)(a — y){x + y — a). Ans. — , 

jut 



5. Find the minimum value of 

x 2 + xy + y 2 — ax— by. Ans. - (ab — a 2 — b 2 ) . 

o 

6. Find the values of x and y that render 

sin x + sin y + cos (x + y) 

a maximum or minimum. A . . . -, 3tt 

Ans. A minimum, when x==y = — ; 

25 

a maximum, when x=y = -. 

6 

7. Find the maximum value of 

(cu; + 6w + c) 2 A 9 . 7 2 , 2 

v — „ - y „ y . ^Ins. or + &- 4- <r. 

^ + z/ 2 + i 



8. Find the maximum value of afyV, subject to the condition 



2x + 3y-t-4z = a Ans. 



9. Find the minimum value of - + - + -, subject to the condition 

a b c 

Xyz = abc. Ans. 3. 



160 DIFFERENTIAL CALCULUS 

10. Divide a into three parts such, that their continued product 
may be the greatest possible. 

Let the parts be x, y, and a — x — y. 

Then u = xy (a — x — y), to be a maximum. 

— = ay-2xy-y 2 = 0, — = ax-x*-2xy = 0. 
dx oy 

These equations give x = y = ^. 

o 

Hence a is divided into equal parts. 

Note. — When, from the nature of the problem, it is evident that there is 
a maximum or minimum, it is often unnecessary to consider the second 
derivatives. 

11. Divide a into three parts, x, y, z, such that x m y n z p may be a 
maximum. 

Ans. * = Z = *-=^ . 

m n p m + n+p 

12. Divide 30 into four parts such that the continued product of 
the first, the square of the second, the cube of the third, and the 
fourth power of the fourth, may be a maximum. 

Ans. 3, 6, 9, 12. 

13. Given the volume a 3 of a rectangular parallel opiped ; find 
when the surface is a minimum. 

Ans. When the parallelopiped is a cube. 

14. An open vessel is to be constructed in the form of a rec- 
tangular parallelopiped, capable of containing 108 cubic inches of 
water. What must be its dimensions to require the least material iu 
construction ? 

Ans. Length and width, 6 in. ; height, 3 in. 

15. Find the coordinates of a point, the sum of the squares of 
whose distances from three given points, 

0*i> 2/i,)> ( x 2 2/ 2 ), (a* 2/s), 



MAXIMA AND MINIMA OF FUNCTIONS 161 

is a minimum. ^ |<*, + *, + **), §<* + *+*>, 

the centre of gravity of the triangle joining the given points. 

16. If x, y, z are the perpendiculars from any point P on the sides 
a, b, c of a triangle of area A, find the minimum value of x 2 + y 2 + z 2 . 

4 A 2 



Ans. 



a 2 + &2 + c 2 



17. Find the volume of the greatest rectangular parallelopiped 
that can be inscribed in the ellipsoid, 

^ + S + S-1 Ans. «* 

a 2 b 2 c 2 3V3 

18. The electric time constant of a cylindrical coil of wire is 

u = mx ^ z , 
ax + by + cz 

where x is the mean radius, y is the difference between the internal 
and external radii, z is the axial length, and m, a, b, c are known con- 
stants. The volume of the coil is nxyz = g. Find the values of x, y, z 
which make u a minimum if the volume of the coil is fixed ; also the 
minimum value of u. 

Ans. ax=by=cz=Jj^- u = ™j[jEL 

\ ?i ' 3 Vabctf 



CHAPTER XIV 

CURVES FOR REFERENCE 

We give in this chapter representations and descriptions of some 
of the curves used as examples in the following chapters. 



RECTANGULAR COORDINATES 
Y 



125. The Cissoid, 



f 



2a — x 



This curve may be constructed from 
the circle ORA (radius a) by drawing 
any oblique line OM, and making 

PM= OR, 

The equation above may be easily 
obtained from this construction. The 
line AM parallel to T is an asymp- 
tote. 

The polar equation of the cissoid is 

r=2a sin 6 tan 0. 



162 




126. The Witch of Agnesi, y = 



CURVES FOR REFERENCE 

8 a 3 



163 



a 2 4-4 a 2 




This curve may be constructed from the circle OBA (radius, a) by 
drawing any abscissa ME, and extending it to P determined by ORN, 
by the construction shown in the figure. 

The equation above may be derived from this construction. The 
axis of X is an asymptote. 



127. The Folium of Descartes, 

b 3 -f- V s — 3 axy = 0. 

The point A, the vertex of 
the loop, is 



fSa S_a\ 
\2> 2/ 



The equation of the asymp- 
tote MN is 

x + y + a = 0. 

The polar equation of the 
folium is 

3 a tan 6 sec 6 

l + tan 8 




a o 
a 





164 



DIFFERENTIAL CALCULUS 



128. The Catenary, 



2/ = 7jV + e «)• 




This is the curve of a cord or chain suspended freely between two 
points. 

129. The Parabola, referred to Tangents at the Extremities of the 

l JL 1 

Latus Rectum, x 2 + y^ = a 2 . 

OL=OL' = a. 
Y 




The line LL' is the latus rectum ; its middle point F, the focus ; 
OFM, the axis of the parabola ; A the middle point of OF, the vertex. 



CURVES FOR REFERENCE 



165 



130. The curve a n ~ l y = .v n , where one coordinate is proportional 
to the nth power of the other, is sometimes called the parabola of 
the nth degree. 

If n = 3, we have the Cubical Parabola, a 2 y = x 5 . 

Y 



If n= -, we have the Semicubical Parabola, 



a 2 y= x 2 , ay 2 = x 3 . 




166 DIFFERENTIAL CALCULUS 

131. The Two-arched Epicycloid. 



x = — -cos 4> — -cos 3 </>, 



y= — -sm<£--sm3<£. 




132. The Hypocycloid of Four Cusps sometimes called the Astroid, 



2. 2. 

a; 3 +y 3 



2 

a 3 - 



This is the curve de- 
scribed by a point P 
in the circumference of 
the circle PR, as it rolls 
within the circumfer- 
ence of the fixed circle 
ABA', whose radius a 
is four times that of 
the former. 

The equation above 
may be given in the 
form 

#=acos 3 <£, 2/=asin 3 <£. 




CURVES FOR REFERENCE 



167 



"/+ (*)*=* *' 



The equation is 
the same as that 
of the ellipse with 
the exponent of 
the second term 
changed from 2 
tof. 




134. The Curve, <ry- — on? — a:' 







POLAR COORDINATES 
135. The Circle, r = a sin $. 



The circle is OPA (diameter, 
a) tangent to the initial line 
OX at the origin 0. 




168 DIFFERENTIAL CALCULUS 

136. " The Spiral of Archimedes, r = aO. 



In this 
curve r is 
proportional 
to 6. Lay- 
ing off 

r = OA, 

when 
= 2tt, 

then 



OP x = \OA, OP 2 = iOA, OP z = \OA, OP 5 = iOA, 
0B = 2 0A, OC = SOA. 

The dotted portion corresponds to negative values of 6. 

137. The Hyperbolic or Reciprocal Spiral, rO 




a. 



In this curve r 
varies inversely as 
0. The line MN 
is an asymptote, 
which the curve 
approaches, as 6 
approaches zero. 

Since r=0 only 
when = oo , it fol- 
lows that an in- 
finite number of 
revolutions are 
necessary to reach the origin. 




CURVES FOR REFERENCE 



169 



r = e 



138. The Logarithmic Spiral 
Starting from A, 

where 0=0 and r=l, 
r increases with 6 : 
but if we suppose 
negative, /• decreases 
as 6 numerically in- 
creases. Since r=0 
only when 0=—x:, 
it follows that an 
infinite number of 
retrograde revolu- 
tions from A is re- 
quired to reach the 
origin 0. 

A property of this spiral is that the radii vectores OP, OPj, OP 2 , 
make a constant angle with the curve. 

139. The Parabola, Origin at Focus, r(l — cos 0) = 2 a. 




The initial line OX is the axis of 
the parabola; the origin is the 
focus ; LL', the latus rectum. 




140. The Parabola, Origin at Vertex (see preceding figure), 

/• sin $ tan 6 = 4a. 
The initial line is the axis AX; the origin is the vertex A. 



170 



DIFFERENTIAL CALCULUS 



141. The Cardioid, r = a(l — cos 6). 

This is the curve described 
by a point P in the circum- 
ference of a circle PA (di- 
ameter, a) as it rolls upon 
an equal fixed circle OA. 

Or it may be constructed 
by drawing through 0, any 
line OB in the circle OA, 
and producing OR to Q and 
Q', making BQ=RQ'=OA. 

The given equation fol- 
lows directly from this con- 
struction. 




142. The Equilateral Hyperbola, r 2 cos 2 



The origin 
is the 
centre, of the 
hyperbola, 
and the in- 
itial line OX 
is the trans- 
verse axis. 

If Or is 
taken as the 
initial line, 
the equation 
of the hyper- 
bola is 

r 2 sin 2 6= a 2 . 




CURVES FOR REFERENCE 171 

143. The Lemniscate referred to OA (see preceding figure), 
r = a 2 cos 2 0. 

This is a curve of two loops like the figure eight. 

It may be defined in connection with the equilateral hyperbola, as 
the locus of P, the foot of a perpendicular from on PQ, any 
tangent to the hyperbola. 

The loops are limited by the asymptotes of the hyperbola, making 

TOX =T'OX= 45°. OA = a. 

The lemniscate has the following property: 

If two points, F and F', called the foci, be taken on the axis, such 



that 



OF= OF' 



V2 



then the product of the distances P'F, P'F', of any point of the 
curve from these fixed points, is constant, and equal to the square 
of OF. 

If J" is taken as the initial line, the equation of the lemniscate is 

r 2 = a 2 sin 2 6. 



144. The Four-leaved Rose, r=a sin 2 0. 




172 



DIFFERENTIAL CALCULUS 



145. The Curve, r = a sin 3 - 

o 




CHAPTER XV 



DIRECTION OF CURVES. TANGENTS AND NORMALS 

We have seen in Art. 17 that the derivative at any point of a 
plane curve is the slope of the curve at that point. We will now con- 
sider some further applications of differentiation to curves. 

146. Subtangent, Subnormal, Intercepts of Tangent. — Let PT be 
the tangent, and PX the normal, to a curve at the point P, whose 
ordinate is y == PM. 
Then MT is called the 
subtangent, and . MN 
the subnormal, corre- 
sponding to the point 
P. 

To find expressions 
for these quantities : 

Let <f> donote the 
angle PTX, the in- 
clination of the tan- 
gent to OX. 

By Art. 17, 




tan PTX = tan <f> = ^-. 
dx 



Subtangent = TM= PM cot PTM= y cot <£ = ^ = y — 



dx 



Subnormal = MN= PM tan MPX= ytsmcf> = y 

Intercept of tangent on OX= OT = OM— TM=x — y 

PS - PM= 



Intercept of tangent on OY= OT' 
But as OT' is negative, we have 
Intercept of tangent on Y= y — x tan cf> = y — x 



dy 

dx 
dx 
dy 

x tan <f> — y. 



dx 



174 



DIFFERENTIAL CALCULUS 



147. Angle of Intersection of Two Curves. Suppose the two curves 
intersect at P. 

Let PT and PT be the 
tangents at P. 

PTX=<j>, PT'X=cf>', 

and let I be the angle 
TPT between the tangents. 
Then I=<f>'-<]> and 

tan/= tan< fr f - tan< ft-. (i) 
l+tan<£'tan<£ v ' 

From the equations of 
the given curves find the 
coordinates of the point of 
intersection P; then using 

these equations separately, find by tan <f> = ^ the values of tan <£ 

dx 

and tan <j>' for the point P. Substituting in (1) gives tan /. 

For example find the angle at which the circle 




T 



dy 



a* + 3^ 13, . . . 

intersects the parabola 
2y 2 = 9x. . . . 



(2) 



(3) 



The intersection P of (2) and (3) is 
found to be (2, 3). 

Differentiating (2), 



^ = -- = -? forP, tan<£ = 
dx y 3 

From (3), 



dy 

dx 4cy 



2 
3* 

9 =?forP, 




tan <f>' = 



Substituting in (1), tan/ 



17 



1=70° 33', 



DIRECTION OF CURVES. TANGENTS AND NORMALS 175 

EXAMPLES 

1. Find the direction at the origin of the curve, 

(a 4 — b i )y = x (x - a) 4 - b 4 x. Ans. 45° with OX. 

What must be the relation between a and b, so that it may be 
parallel to OX at the point x = 2 a ? • Arts. 3a 2 = b 2 . 

2. Find the poiuts of contact of the two tangents to the curve, 

6 y = 2 X s + 9 x 2 - 12 aj + 2, 
parallel to the tangent at the origin to 
the curve, y 2 + ay = 2 ax. Ans. f 1, - V ( — 4, 11)- 

3. Find the subtangents and subnormals in the parabolas, 

7f = 4tax, and x 2 = 4:<iy. 

Ans. Subtangents, 2a?, -; subnormals, 2 a, — — 

4. Find the subtangent and subnormal in the cissoid (Art. 125), 

y 1 = - 1 , at the point (a, a). Ans. -, 2 a. 

Li (X X Li 

5. Show that the sum of the intercepts of the tangent to the 

iii 
parabola (Art. 129), # 2 + y 2 = a 2 , is equal to a. 

6. Show that the area of the triangle intercepted from the co- 
ordinate axes by the tangent to the hyperbola, 

2 xy = a 2 , is equal to a 2 . 

7. Show that the part of the tangent to the hypocycloid (Art. 132), 

2 2 2 

— a 3 , intercepted between the coordinate axes, is equal to a, 

8. At what angle do the parabolas, y 2 = ax and x 2 = 8 ay intersect? 

3 
An s. At (0, 0), 90°; at another point, tan -1 -• 



176 DIFFERENTIAL CALCULUS 

9. At what angle does the circle, x 2 + y 2 = 5 x, intersect the' curve, 
3 y = 7 x 3 — 1, at their common point (1, 2) ? Ans. 45°. 

10. Show that the ellipse and hyperbola, 

' 7 + 2 ' 3 2~ ' 
intersect at right angles. 

11. Find the angle of intersection of the circles, 

x * + y 2-. x + 3y + 2 = 0, x 2 + if-2y = 9. Ans. tan" 1 \ 

12. Show that the parabola and ellipse, 

y 2 = ax, 2 x 2 + y 2 = 6 2 , 
intersect at right angles. 

13. Show that the parabolas, 

y 2 = 2 ax + a 2 , and # 2 = 2 % + 6 2 , 
intersect at an angle of 45°. 

14. Find the angle of intersection of the parabola, 

x 2 = 4 ay, and the witch (Art. 126), y = - 8 



^bis. tan- 1 3=71°34'. 

15. Find the angle of intersection between the parabola, 
y 2 = 4 a#, and its evolute, 27 a?/ 2 = 4 (x — 2 a) 3 . (See Fig., Art. 167.) 

-4ns. tan -1 

V2 

148. Equations of the Tangent and Normal. Having given the 
equation of a curve y=f(x), let it be required to find the equation 
of a straight line tangent to it at a given point. 



DIRECTION OF CURVES. TANGENTS AND NORMALS 177 

Let (V. y 1 ) he the given point of contact. Then the equation of a 
straight line through this point is 

y-y' = m{x-x'), (1) 

in which x and y are the variable coordinates of any point in the 
straight line ; and m, the tangent of its inclination to the axis of X. 
But since the line is to be tangent to the given curve, we must have, 
by Art. 17, 



m = tan (f> 



dx' 



( -^ being derived from the equation of the given curve y =f(x), 
and applied to the point of contact (»', ?/'). 

If we denote this by — , we have, substituting m = -^- in 

ii m\ dx' dx' 

equation (1), 

y-y'=%(*-*), (2) 

for the equation of the required tangent. 

Since the normal is a line through (a?', y') perpendicular to the 
tangent, we have for its equation 

r-f—^C— 0— g<— ■> (3) 

dx' 

For example, find the equations of the tangent and normal to the 
circle x 2 + if = a 2 , at the point (x', y'). 

Here, by differentiating ar + ?/ 2 = cr, we find 

— = — -, from which -&- = — — . 
tw y dx' y' 



Substituting in (2), we have 



x' 



y-y' = --X x ~ x % 
y ■ 

as the equation of the required tangent. 



178 DIFFERENTIAL CALCULUS 

It may be simplified as follows : 

yy'-y' 2 = -xx' + x' 2 , 

xx' -\-yy'=x' 2 + y' 2 = a 2 . 
The equation of the normal to the circle is found from (3) to be 

y-y'=-,(x-x'), 

X 

which reduces to x'y = y'x. 

EXAMPLES 

Find the equations of the tangent and normal to each of the three 
following curves at the point (x' } y') : 

1 . The parabola, y 2 = 4 ax. 

Ans. yy' =2 a{x + x'), 2a(y — y')+y'(x — x r ) = 0. 

2. The ellipse, ^ + ^ = 1. 



a 2 b 2 



Ans. ^ + WL = 1, b 2 x'(y - y')= a 2 y'(x - a?'). 



3. The equilateral hyperbola, 2xy = a 2 . 

Ans. xy' + yx' = a 2 , y'(y — y') = x'(x — x'). 

4. Find the equation of the tangent at the point (x',y') to the 
ellipse, 3x 2 -4:xy + 2y 2 + 2x = 2. 

Ans. 3 xx' + 2yy'—2 (x'y + y'x) + x + x' = 2. 

5. Find the equations of tangent and normal at the point (V, y') 
to the curve, x 5 — a s y 2 . 

Ans. ^-^ = 3, 2xx' + 5yy' = 2x' 2 + 5y' 2 . 
x y 



DIRECTION OF CURVES. TANGENTS AND NORMALS 179 

ar 3 

6. In the cissoid (Art. 125), y 2 = , find the equations of the 

2a — x 

tangent and normal at the points whose abscissa is a. 

Ans. At (a, a), y = 2 x — a, 2 y -f- x = 3 a. 
t At (a, — a), y + 2x = a, 2y — x — 3a. 

So 3 

7. In the witch (Art. 126), y = — , find the equations of 

4r + x- 

the tangent and normal at the point whose abscissa is 2 a. 

Ans. x + 2y = ka, y = 2 x — 3 a. 

8. Find the equation of the tangent at the point (x', y') to the 

curve, X s y + xy 2 = a 3 . 

Ans. xy\2 x' + y') + yx'(2 y' + x') = 3 a 3 . 

Find the equations of tangent and normal to the three following 
curves : 

9. x'+f =3 axy (Avt. 127), at /^, 2^). Ans. x+y=3a, x=y. 

10. :c + y = 2e x - y , at (1, 1). Ans. 3y = x-{-2, 3^ + y = 4. 

11. fjY + (llX= 2, at (a, b). Ans. - + U. = 2, ax -by = a 2 - b 2 . 
\aj \bj a b 

12. Find the equations of the two tangents to the circle, 

x 2 + y 2 — 3y = 14:, parallel to the line, 7y = 4:X + l. 

Ans. 7 2/ = 4# + 43, ly = 4 x — 22. 

13. Find the equations of the two normals to the hyperbola, 

4 x 2 - 9 y 2 + 36 = 0, parallel to the line, 2 y + 5 x = 0. 

^[?is. 8 y + 20 a; =±65. 

149. Asymptotes.* "When the tangent to a curve approaches a 
limiting position, as the distance of the point of contact from the 
origin is indefinitely increased, this limiting position is called an 

* The limits of this work allow only a brief notice of this subject. 



180 DIFFERENTIAL CALCULUS 

asymptote. In other words, an asymptote is a tangent which passes 
within a finite distance of the origin, although its point of contact is 
at an infinite distance. 

We have found in Art. 146, for the intercepts of the tangent on the 
coordinate axes, 

Intercept on OX=x — y — , Intercept on OY= y — x c -^-. 
dy dx 

If either of these intercepts is finite for x = oo, or y = oo, the cor- 
responding tangent will be an asymptote. 

The equation of this asymptote may be obtained from its two 

intercepts, or from one intercept and the limiting value of -^. 

dx 

Let us investigate the conic sections with reference to asymptotes. 

(1) The parabola, y 2 = 4 ax, ^ = ^. 

dx y 

dx ii~ 

Intercept on OX — x — y— — x — ^— = — x, 
dy 2 a 

t x. , ^ Tr dy 2 ax y 

Intercept on OT = y — x— ' = y = ^ . 

dx y 2 

When x = oo, y=ao, and both intercepts are also infinite. 
Hence the parabola has no asymptote. 

(2) The hyperbola, .^-«j = l, &=.$£. 

a 2 b 2 dx a 2 y 






Intercept on OX = — , Intercept on OY = . 

x y 

These intercepts are both zero when x = oo, and there is an 
asymptote passing through the origin. To find its equation, it is 

necessary to find the limiting value of -^, when x = oo. 



Hence 



dy _ b 2 x _ bx 


-± b - 1 




dx a 2 y a V^2 _ 


-« 2 \/i- 


a 2 

X 2 


dx a' 


when x =oo. 


9 



DIRECTION OF CURVES. TANGENTS AND NORMALS 181 

There are then two asymptotes, whose equations are 

, b 
y = ± -x. 
a 

(3) The ellipse, having no infinite branches, can have no 
asymptote. 

150. Asymptotes Parallel to the Coordinate Axes. When, in the 

equation of the curve, x = oo gives a finite value of y, as y = a, then 
y = a is the equation of an asymptote parallel to OX. 

And when y = oo gives x = a, then x = a is an asymptote parallel 
to OY. 



151. Asymptotes by Expansion. Frequently an asymptote may 
be determined by solving the equation of the curve for x or y, and 
expanding the second member. 

For example, to find the asymptotes of the hyperbola 

y = ± V - ^= ± *?(l - f)i= ±*(l*-. 

a a\ arj a\ 2x 2 

As x increases indefinitely, the curve approaches the lines 
y = ± — , the asymptotes. 



EXAMPLES 

Investigate the following curves with reference to asymptotes : 

x 3 

1. ? = 32 , 3 a » - Asymptote, ?/= z. 

2. y 3 = 6x 2 — or 5 . Asymptote, a; + y = 2. 

3. The cissoid (Art. 125) y 2 = ^ Asymptote, a; = 2a. 



182 



DIFFERENTIAL CALCULUS 



4. a? 3 + ?/ 3 = a 3 . Asymptote, x + y = 0. 

5. (a? — 2 a)?/ 2 = x 3 — a 3 . Asymptotes, # = 2 a, x + a = ± ?/. 

6. X s -f 2/ 3 = 3 cm/ (Art. 127). Asymptote, x + ?/ + a = 0. 

(Substitute ?/ = vx in the given equation and in the expressions 
for the intercepts.) 

152. Direction of Curve. Polar Coordinates. 



In this case the angle 
OPT between the tangent 
and the radius vector may 
be most readily obtained. 
Denote this angle by if/. 
Let r, 0, be the coordi- 
nates of P; r+Ar,0+A0, 
the coordinates of Q. 
Draw PR perpendicular 
to OQ. 



Then tanPQP = |^ = 

RQ r + A> 




— rcos A0 

sin A0 

A0 



Ar + 2rsin 2 ^ 



. A0 

sin — 

Ar • A6» 2 

— - -f r sin — — 

A0 2 A0 



Now let A0 approach zero ; the point Q approaches P, and the 
angle PQR approaches its limit if/. 



Hence 



tan if/ = Lim A =o tan PQR = 



d0 



(1) 



The inclination <f> of the tangent to OX may be found by 

<f> = t+0 (2) 



DIRECTION OF CURVES. TANGENTS AND NORMALS 183 
153. Polar Subtangent and Subnormal. 



If through O, NT be drawn per- 
pendicular to OP, OT is called the 
polar subtangent, and ON the polar 
subnormal, corresponding to the 
point P. 

OT=OP tan OPT; that is, 

r 2 
Polar subtangent = r tan \b = — . 

a> 

dO 




ON= OP cot PNO; that is, 

dr 

Polar subnormal = r cot xb = — • 

Y dO 



154. Angle of Intersection. Suppose the two curves intersect at P, 
and have the tangents PT and PT\ 
OPT=xj / , OPT' = x\,'. 

Then the angle of intersection, 



and 



tan I- tan ^-tani/r 
1 + tan i// tan ^ 



a) 




By this formula the angle of inter- 
section may be found in polar coordi- 
nates, in the same way as by (1), Art. 147, in rectangular coordinates. 

For example, find the angle of intersection between the curves 

r = asin20, (2) 

and r = a cos 20. (3) 

From (2) and (3) we have for the intersection 

tan20 = l. 



184 DIFFERENTIAL CALCULUS 

From (2), tan if/' = - tan 2 = -, for the intersection. 

-i Z 

From (3), tan if/ = — - cot 2 = — -, for the intersection. 

Z Z 

Substituting in (1), tan I — - . 

o 

The curves are that in Art. 144, and the same curve revolved 45 c 
about the origin. 



EXAMPLES 

1. In the circle (Art. 135), r — a sin 6, find if/ and <£. 

Ans. if/ = 6, and $ = 20. 

2. In the logarithmic spiral (Art. 138), r = e ad , show that \J/ is 
constant. 

3. In the spiral of Archimedes (Art. 136), r = ad, show that 
tan \j/=0; thence find the values of ^, when = 2 -k and 4 -k. 

Ans. 80° 57' and 85° 27'. 
Also show that the polar subnormal is constant. 

4. The equation of the lemniscate (Art. 143) referred to a tangent 
at its center is r 2 = a 2 sin 2 0. Find if/, <f>, and the polar subtangent. 

Arvs. if/ = 2 6; <£ = 30; subtangent = a tan 2 Vsin 2 0. 

5. In the cardioid (Art. 141), r = a (1 — cos 0), find <£, ^, and the 
polar subtangent. 

Ans. <f> = — ; if/ = - ; subtangent = 2 a tan - sin 2 — 

z z z z 

6. Find the area of the circumscribed square of the preceding 

cardioid, formed by tangents inclined 45° to the axis. 

27 — 

AnS ' f^ 2 + V3)a 2 . 



DIRECTION OF CURVES. TANGENTS AND NORMALS 185 

7. In the folium of Descartes (Art. 127), r= 3atan ^ sec ^ 



show that tan <£ = 



+ 

tan 4 - 2 tail 6 
2tan 3 0-l 



8. Find the area of the square circumscribed about the loop of 
the folium of the preceding example. 

Ans. 2 a/2 a 2 . 

9. Show that the spiral of Archimedes (Art. 136), r = ad, and the 
reciprocal spiral (Art. 137), rO = a, intersect at right angles. 

10. Show that the cardioids (Art. 141), 

r = a (1 — cos 0), r = b (1 + sin 6), 
intersect at an angle of 45°. 

11. Show that the parabolas (Art. 139), 

r =: m sec- -, r = n cosec- - 

2' 2' 

intersect at right angles. 

12. Find the angle of the intersection between the circle (Art. 135), 
r = a sin 0, and the curve (Art. 144), r = a sin 2 6. 

Ans. At origin 0° ; at two other points, tan -1 3 V3 = 79° 6'. 

13. Find the angle of intersection between the circle (Art. 135), 
r = 2 a cos 6, and the cissoid (Art. 125), r = 2 a sin tan 0. 

Ans. tan -1 2. 

14. At v% hat angle does the straight line, r cos 6 = 2a, intersect the 
circle (Art. 135), r = 5 a sin 6 ? ^ tan _, 3 

4 

15. Show that the equilateral hyperbolas (Art. 142), r 2 sin 2 6 = a 2 , 
/-cos 26 = b 2 , intersect at right angles. 



186 



DIFFERENTIAL CALCULUS 



16. Find the angle of intersection between the circles 
r = a sin + b cos 0, r = acos$-\-b sin 0. 

Ans. tan" 



2ab 



17. Find the angle of intersection between the lemniscate (Art.143), 
r 2 = a 2 sin 2 0, and the equilateral hyperbola (Art. 142), f^sin 2 6 — b 2 . 

Ans. 2 sin -1 — 



155. Derivative of an Arc. Rectangular Coordinates. Let s denote 
the length of the arc of the curve measured from any fixed point of it. 



Then 
We have 



arc AP, As = arc PQ. 



sec QPR 



PQ 
PR 



Now suppose Ax to approach zero, and consequently the point Q 
to approach P. 
Then 

Lim sec QPR= sec TPR= sec <£. 

PQ = PQ arcPQ 
PR arcPQ* PR ' 

since 

Lim _p_e_ =1 , 

arcPQ 

Lim^=l4m^§ 
P# PR 

T . As c?s 
Ax dx 

ds 




Hence 



sec<£ = 



dx' 



therefore |= VH^-V^g) '■ 



(1) 






DIRECTION OF CURVES. TANGENTS AND NORMALS 187 



It is evident also that 

SlU(i = -, COS 6= . . (2) 

ds ds 

It may be noticed that these 
relations (1) and (2) are cor- 
rectly represented by a right 
triangle, whose hypothennse is 
ds, sides dx and dy, and angle 
at the base <j>. 




Here 



ds= ^(dxf+(dyy, 



or 



dx V [dxj 



156. Derivative of an Arc. Polar Coordinates. From the figure of 
Art. 152, we have, as A0 approaches zero, 

sec \b = Lim sec FOR = Lim — -^ = Lim ar ° ^ = Lim — J. 
r ^ RQ RQ RQ 



As 



As 



As 
A0 



*' + »'-»¥ A,,.^ . A. 



A0 



AF sm T 



Hence 



ds 

. T • As d9 ds 

sec ^ = Lim = — = — , 

Y RQ dr dr 



dO 



-^ r = Vl+tan 2 i/> 



V' + <|J, 



<l»_'l*<]r_ . . /'*' 



Vctf 



(i) 

(2) 
(3) 



188 



DIFFERENTIAL CALCULUS 



It may be noticed that these relations (1), (2), and (3), are cor- 
rectly represented by a right triangle, whose hypothenuse is ds, sides 
dr and rdd, and angle between dr and ds, \p. 




Here ds = V (drf + (r ddf, 

and thence ^ = Ji + W™Y, or * 



=V r2 



■I I)- 



CHAPTER XVI 
DIRECTION OF CURVATURE. POINTS OF INFLEXION 

157. Concave Upwards or Downwards. A curve is said to be con- 
cave upwards at a point P, when in the immediate neighborhood of 
P it lies wholly above the tangent at P, as in the first figure below. 
Similarly, it is said to be concave downwards, when in the immediate 
neighborhood of P it lies wholly below the tangent at P, as in the 
second figure below. 

It will now be shown that when the equation of the curve is in 
rectangular coordinates, the curve is concave upwards or downwards, 

d 2 v 
according as —- is positive or negative. 
ax~ 




Suppose -^ > 0, that is, — ( — ) > 0; in other words, the derivative 
oar dx\dxj 

of the slope is positive. 

Then by Art. 21 the slope increases as x increases. 

This case is illustrated in the first figure above, where the slope 
evidently increases as we pass from P Y to P 2 . The curve is then con- 
upwards. 

But if --» < 0, it follows that the slope decreases as x increases. 

189 



190 



DIFFERENTIAL CALCULUS 



We then have the case of the second figure, where the slope de- 
creases as we pass from P 1 to P 2 . The curve is then concave down- 
wards. 



158. A Point of Inflexion is a point P where — ^ changes sign, 

dx 2 

the curve being concave upwards on one side of this point, and con- 
cave downwards on the other. 



dx 2 



This can occur, provided -^ and 

dx 



are continuous, only when 



^ = 
dx 2 



(1) 



But if -^ and — \ are infinite, we 
dx dx 2 

may have a point of inflexion 

when J = oc 

dx 2 




X* 



It is evident that the tangent at a point of inflexion crosses the 
curve at that point. 

For example, find the point of inflexion of the curve 

2y = 2-8x + 6x 2 -x 3 . 

d 2 y_ 



Here 



dx 2 



3(2-a>). 



Putting this equal to zero, we have for the required point of in- 
flexion, x = 2. If x<2, ^ o >0; and if a>>2,^<0. 
' dx 2 ' dx 2 

Hence the curve is concave upwards on the left, and concave down- 
wards on the right, of the point of inflexion. 



DIRECTION OF CURVATURE. POINTS OF INFLEXION 191 

EXAM PLES 

Find the points of inflexion and the direction of curvature of the 
five following curves : 

1. y=(r-iy. 

Ans. x= ± — -; concave downwards between these points, con- 
V3 

cave upwards elsewhere. 

2. y = x* - 16 x" + 42 x 2 - 28 x. 

^Lns. 05 = 1 and x= 7; concave downwards between these points, 
concave upwards elsewhere. 

3. a 4 y = x (x — a) 4 + a A x. 

r> a 

Ans. x=^; concave downwards on the left of this point, con- 
o 

cave upwards on the right. 



4. The witch (Art. 126), y 



Ans. ( ± — *-, — ) : concave downwards between these points, 
concave upwards outside of them. 



5. The curve, y 



a? 



a? + 3 a* 



Ans. [—3a, -j, (0,0), (3 a, — ); concave upwards on the 

left of first point, downwards between first and second, 
upwards between second and third, and downwards on the 
right of third point. 



192 DIFFERENTIAL CALCULUS 



Find the points of inflexion of the following curves: 

4:X 

x> + ± 



4t x m 

6 - y = „a ■ a ' Ans. x = and ± 2 V3. 



7. y = T «*. Ans. x = -2a. 

(x — a) 2 

8.i/ = (# 2 -f a?) e -x . Ans. x = and a; = 3. 

9. y = e-~-e-K Ans. x = ^2S^zMH. 

a — b 

10 . gy + g)U. ^ „. ± -. 



11. aV = a¥ - » 6 (Art. 134). Ans. x=±% ^21 - 3 V33. 

6 



CHAPTER XVII 

CURVATURE. RADIUS OF CURVATURE. EVOLUTE AND 

INVOLUTE 

159. Curvature. If a point moves in a straight line, the direc- 
tion of its motion is the same at every point of its course, but if its 
path is a curved line, there is a continual change of direction as it 
moves along the curve. This change of direction is called curvature. 

We have seen in the preceding chapter that the sign of the second 
derivative shows which way the curve bends. We shall now find 
that the first and second derivatives give an exact measure of the 
curvature. 

The direction at any point being the same as that of the tangent 
at that point, the curvature may be measured by comparing the 
linear motion of the point with the simultaneous angular motion of 
the tangent. 

160- Uniform Curvature. The curvature is uniform when, as the 
point moves over equal arcs, the tangent turns through equal angles. 
The only curve of uniform curvature is the circle. Here the meas- 
ure of curvature is the ratio between the angle described by the tan- 
gent and the arc described by the point of contact. In other words, 
it is the angle described by the tangent while the point describes a unit 
of arc. 

Suppose the point Pto move in the circle AQ. 

Let s denote its distance AP from some initial position A } and 
<j> the angle PTXmade by the tangent PT with OX. 

Then as the point moves from P to Q, s is increased by PQ = As, 
and <f> by the angle QRK= A<£. 

As the point describes the arc As, the tangent turns through the 
angle A(£. 



194 



DIFFERENTIAL CALCULUS 



The curvature, being uni- 
form, is then equal to — • 
' ^ As 

If we draw the radii CP, 
CQ, and let r denote the 
radius, then 

angle PCQ = QBK= A</>. 

But 
arc PQ = CP (angle PCQ) ; 

that is, 

As = rA<f>, — ^ = — 
As r 




Hence the curvature of a circle is the reciprocal of its radius. 
For example, suppose the radius of a circle to be 50 feet. 



Then its curvature is 



A<j> = 1 
As ~ 50' 



where A<f> is in circular measure, and As in feet. 

In other words, for every foot of arc, the change of direction is 

— in circular measure = 1° 8' 45". 
50 



161. Variable Curvature. For all curves except the circle the 
curvature varies as we move along the curve. In moving over the 

arc As, — * is the mean curvature throughout the arc. The curva- 



As 



ture at the beginning of this arc is more nearly equal to — *, the 
shorter we take As. 

Hence the curvature at any point of a curve is equal to 

-L<ini As= o - — — -7- 
As as 



CURVATURE. RADIUS OF CURVATURE 



195 



162. Circle of Curvature. A circle tangent to a curve at any point, 
having its concavity turned in the same direction, and having the 
same curvature as that of the curve at that point, is called the circle 
of curvature ; its radius, the radius of curvature; and its centre, the 
centre of curvature. 

The figure shows the circle of curvature MPN for the point P of the 
ellipse. C is the centre of curvature, and CP the radius of curvature. 

It is to be noticed 
that the circle of curv- 
ature crosses the curve 
at P. This can be 
easily proved. 

At P the circle and 
ellipse have the same 
curvature, but as we 
go towards P 2 , the 
curvature of the ellipse 
increases, while that of 
the circle continues the same. 

Hence on the right of P the circle is outside of the ellipse. 

Moving from P to P 2 , the curvature of the ellipse decreases, and 
therefore on the left of P the circle is inside of the ellipse. 

So in general the circle of curvature crosses the curve at the point of 
contact. 







B 


n>\ 


i! VL 




















/ p!v 











_j^C^ 


A J \ 




M 













N 




196 DIFFERENTIAL CALCULUS 

The only exceptions to this rule are at points of maximum and 
minimum curvature, as the vertices A and B of the ellipse. 

As we move from A along the curve in either direction, the curva- 
ture of the ellipse decreases ; hence the circle of curvature at A lies 
entirely within the ellipse. 

Similarly it appears that the circle of curvature at B lies entirely 
without the ellipse. 

163. Radius of Curvature. The curvature of the circle of curva- 
ture being that of the given curve, is equal to -^ (Art. 161). If we 

denote the radius of curvature by p, then by Art. 160, 

ds m 

p= ^ (1) 

ds 

To obtain p in terms of x and y, we may write (1), p = — = — - • 

dcf> d<f> 



dx 



From (1) Art. 155, ^ = Jl + f§Y 

dx \ ^ \dx) 



Also, tand> = ^, <f> = tan- 1 ^\ 

dx \dxj 



Differentiating, ^> = ^— - (2) 



dcf> dx' 2 




dx 1+ (^y 

\dxj 


Mm 


3 

2 


d 2 y 
dx 2 





Hence P= ^ ^ 



CURVATURE. RADIUS OF CURVATURE 197 

It is to be uoticed that p is always to be considered positive ; that 

the sign of |~1 + f^t 

By interchanging x and y, we have 



is taken the same as that of — . 

dor 



Kf)t 



dtf 

which is sometimes the more convenient expression. 

As an example, find the radius of curvature of the semicubical 
parabola ay 2 = x 3 (Art. 130). 

Differentiating, *_**, ^ = _J 

Substituting in (3), we find 

a£(4 a +9x)i 

P== Wa 



164. Radius of Curvature in Polar Coordinates. Eesuming (1), 
Art. 163, p = —q, let us express p in terms of r and 6. 

ds_ 
d* do 



We may write p 

J H d<f> d± 

dO 



From (3), Art. 156, ds = J? 
0.6 \ 



ds I o , fdr^ 



From (2), Art. 152, 



* *' 06 dO 



198 



DIFFERENTIAL CALCULUS 



From (1), Art. 152, 



T f T 

tan \b-=—, il/ = tan _1 
T dr r 



Differentiating, 



Substituting, 



Hence 



dO 



dr 
ld$) 



'dr\ _ d 2 r 

d&__ \dd) r d& 



dcf> 
dO 



d$ J 

.2,2^— Y-r — 
\d0)__ dO 2 

dr 

dO 



P = 



.2,[dr 
\dO. 



^y d0 2 



(1) 



EXAMPLES 

Find the radius of curvature of the following curves : 

1. y = (x-l) 2 (x-2), at (1, 0) and (2, 0). Ana. p = \ and -L 

2 Y2 

2. ?/ = log x, when a; = f . Ans. p = 2||. 

3. The cubical parabola (Art. 130), ct 2 y = # 3 . Ans. p — T 4 . 



4. The parabola, y 2 = 4ax. 

Find the point of the parabola where p = 54 a. 



Ans. x = 8a. 



( x 2 -4- y 2 ) 2 
5 The equilateral hyperbola, 2xy = a 2 . Ans. p = ± — 2 • 



CURVATURE. RADIUS OF CURVATURE 199 

6. The ellipse, t. + 1 = 1. Am. P = W + **)* . 

ar b- a 4 6 4 

What are the values of p at the extremities of the axes ? 

Ans. - and £L 2 . 
a b 

7. Show that the radius of curvature of the curve, 

x* + y 2 + 10 x — 4 y + 20 = is constant, and equal to 3. 

Find the radius of curvature of the following curves : 

8. sH-log(l-x*) = 0. Ans. P = <■}+% 

9. sin y = e z . Ans. p = e~ x . 

XT ° 

10. The catenary (Art. 128), y = *(e° + e~"«). ^%s. p = £. 

11. The hypocycloid (Art. 132), a?« + y* = a*. ^^- p = 3 (aa;y)*- 

12. The curve aY = aV— a? (Art. 133), at the points (0, 0) and 
(a, 0). Ans. p = - and p = a. 

13. The cycloid, x = a(0 — sm0), y = a(l — cos0). 

Ans. p = 4asin -' 

14. Show that the radius of curvature of the logarithmic spiral 
(Art. 138), r = e ae , is proportional to r. p = r Vl + a 1 . 

15. Show that the radius of curvature of the curve, 

r = a sin 6 4- b cos 0, is constant. p = - Va 2 + 6 2 . 



16. The spiral of Archimedes (Art. 136), r = aO. 

(r^-Mry 
r l + 2 a 2 



Ans. p 



200 DIFFERENTIAL CALCULUS 

17. The cardioid (Art. 141), r = a (1 - cos 6). 



Ans. p 2 = - ar. 



18. The curve, r = a sin 3 - (Art. 145). 



Ans. p=-c&sin 2 -' 
H 4 3 



19 



. The parabola (Art. 139), r — a sec 2 - • Ans. p = 2a sec s 



20. The lemniscate (Art, 143), r 2 = a 2 cos2 



^4ws. o = 



3r 



165. Coordinates of the Centre of Curvature. Let x, y be the co- 
ordinates of P, any point of the curve AB, and C the corresponding 
centre of curvature. CP is 
then the radius of curvature, 
and is normal to the curve. 

Draw also the tangent PT. 

Then CP = p ; 
angle PCR = PTX= <f>. 

Let a, /?, be the coordinates 
of C. OL = OM- RP, 
LC=MP+RC; 
that is, a =x — p sin <£, 

P = y + P cos'<f>. (1) 
To express « and /? in terms of x and y, we have, by (2), Art. 155, 
and (1), (2), Art. 163, 




p sin</> 



pCOS cji 



dy 

ds dy _ dy _dy dx _ cto 
dcf> ds d<j> dx d<fi 



dx dx \dxj 



1 + 



dy s 

dx 



d 2 y 
dx 2 



Hence 



dy 

dx 



1 + 

~¥y 

dx 2 



d 2 y~ 
dx 2 



£=2/ + 



d*y 

dx 2 



(2) 



CURVATURE. RADIUS OF CURVATURE 



201 



166. Evolute and Involute. Every point of a curve AB has a 



Thus, P x , P>, P,, etc., have for 



corresponding centre of curvature, 
their respective centres of 
curvature C 1} C 2 , C 3 , etc, The 
curve UK, which is the locus 
of the centres of curvature, is 
called the evolute of AB. To 
express the inverse relation, 
AB is called the involute 
of HK. 



167. To find the Equation 
of the Evolute of a Given Curve. 
By (2), Art. 165, a and /?, the 

coordinates of any point of the required evolute, may be expressed 
in terms of x and y, the coordinates of any point of the given curve. 
These two equations, together with that of the given curve, furnish 
three equations between a, /3, x, and y, from which, if x and y are 
eliminated, we obtain a relation between a and /3, which is the 
equation of the required evolute. 




For example, find the equation of the evolute of the parabola 



Here 



y- — 4 ax. 

-J- — a?x ?, -^ = crx ' 

dx ' dx 2 2 



Substituting in (2), Art. 165, we have 
a = 3 x + 2 a, j3 = 



Eliminating x, we have for the equation of the evolute, 
«/3 ? = i(«-2«) 3 . 

This curve is the semicubical parabola (Art. 130). The figure 
shows its form and position. F is the focus of the given parabola. 

OC=2a = 2 0F. 



202 



DIFFERENTIAL CALCULUS 



As another example, let us 
find the equation of the e volute Y 
of the ellipse, 9 2 

-+2- =1. 

a 2 b 2 

dy = _b 2 x d 2 y = __&*_ # 
dx a 2 y dx 2 a 2 y 

(Art. 66) 

Substituting in (2), Art. 165, 

To eliminate a; and y be- 
tween these equations and that 
of the ellipse, we find 



x 3 



aa 



y s 



M 



a 3 a 2 - b* b s 



a 2 - b 2 ' 




x 2 ,V_ (q«)* + (&£)* __■■ 



" u (a 2 - b 2 y 

giving, for the equation of the evolute, 

(aa)$+(60)*=(tt 2 ^& 2 )"*. 

The evolute is EF'EFE. E is centre of curvature for A\ C for 
P; Ffor JB; ^' for ^ f ; i^ 7 ' for B'. 

In the figure F and _F' are outside the ellipse, but if the eccentric- 
ity is decreased, so that a < b V2, these points fall within the ellipse. 



168. Properties of the Involute and Evolute. Let us return to the 
equations, (1), Art. 165, 

a = x — p sin <f>, 

£ = y -h p cos <f>. 



CURVATURE. RADIUS OF CURVATURE 



203 



Differentiating 


with 


respect to s, 




da _ dx dp 

ds ds ds 




sin (p — p cos <f> — * 
ds 


a) 


ds ds ds 





coscp-psincp^-. (2) 



Substituting in (1), 

P = ^- and cos 4,= ^, 
d<f> r ds 

(Art. 155), two terms 
cancel each other, 
giving 



Similarly in (2), p=— and sin <f> 

d<p 

d^ dp 

ds ds 




dy 
ds 



(Art. 155), giving 



(4) 



(5) 



Dividing (4) by (3), 3£= L_ 

da tan <£ 

But JS i s the slope of the tangent to the evolute at any point d, 
(see fig., Art. 166), and tan </> the slope of the tangent to the involute 
at the corresponding point P v Since by (5) one is minus the recip- 
rocal of the other, these tangents are perpendicular to each other. 
In other words, a tangent to the evolute at any point C\ is C^, the 
normal to the involute at P v 



204 DIFFERENTIAL CALCULUS 

169. Again, from (3) and (4), Art. 168, 



daV + M?Y fdp\* 0T (d*y = (dp 
dsj \dsj \dsj ' \ds J 



fdp\ 

\dsj' 



where s r denotes the length of the arc of the evolute m&asured from 
a fixed point. Hence, 

ds ds ds\ J 
Hence, s' ± p = a constant, (1) 

since, if a derivative is always zero, the function can neither increase 
nor decrease, but is constant. 
It follows from (1) that 

A 0' ± p) = 0, As' = ± Ap. 

That is, the difference between any two radii of curvature P X C X , 
P 3 (7 3 , is equal to the corresponding included arc of the evolute C X C Z . 

170. From the two properties of Arts. 168 and 169, it follows that 
the involute AB may be described by the end of a string unwound 
from the evolute UK. From this property the word evolute is 
derived. 

It will be noticed that a curve has only one evolute, but an infinite 
number of involutes, as may be seen by varying the length of the 
string which is unwound. 

EXAMPLES 

1. Find the coordinates of the centre of curvature of the cubical 
parabola (Art. 130), ahj = x\ a < lg x < ^ _ 9 ^ 

6 a 2 x 2 a 4 

2. Find the coordinates of centre of curvature of the semicubical 
parabola (Art. 130), ay 2 = x\ 

Ans. . = --!£ ^(* + |)^i 



CURVATURE. RADIUS OF CURVATURE 205 



3. Find the coordinates of the centre of curvature of the catenary 

(Art. 128), y = | ( e -» + <T°). 

^ins. a = x — 
a 



Ans. a = x-^^/ y 2_ a 2 j p == 2y. 



4. Show that in the parabola (Art. 129), x? + y* = a 2 , we have the 
relation a + /3 — 3 (as + y). 

5. Find the coordinates of the centre of curvature, and the equa- 
tion of the evolute, of the hypoeycloid (Art. 132), X s -j-y 1 = d s . 

Ans. a=a +3x*y* } (3 = y + 3 x f y s , 

(« + j8)* + (« - P)*=2a\. 

6. Given the equation of the equilateral hyperbola 2 xy — a 2 , 

show that « + j8= (y+ g a?)3 , «-P= ( y ~^\ 

a- ar 

Thence derive the equation of the evolute, 

(«+/^-(a-/?)3 = 2al 

7. Find the equation of the evolute of the cissoid (Art. 125), 

f = - . Ans. 4096 a s a + 1152 a 2 £ 2 + 27 ^ = 0. 

_ (X — x 



CHAPTER XVIII 

ORDER OF CONTACT. OSCULATING CIRCLE 

171. .Order of Contact. Let us consider two curves whose equa- 
tions are 

■y = <f>(x) and y = if/(x). 

If for a definite value a, of x, the value of y is the same for both 
curves, that is, if 

the curves have 'a common 
point P. 

If, moreover, for x = a, the 

value of — also is the same for 
dx , 

both curves, that is, if 
<f}(a) = if/(a) and <£'(a) = ^'(a), 

the curves have a common tangent at P. 

The curves are then said to have a contact of the^rs^ order. 

If besides, for x = a, the values of — ^ are the same for both 
curves, that is, if x 

<£(a) =ip(a), <£'(a) = i//(«), and <f>"(a) = if/" (a), 

the curves have contact of the second order. 

In general, the conditions for a contact of the nth order at the 
point x = a, are 

*(a) = ^(a), <£'(a) = f(a), *"(a) = f (a), ••, <£»(a) = ^(a), 

and <£ n+1 (a)9^ w+1 <>)- 

206 




ORDER OF CONTACT. OSCULATING CIRCLE 



207 



In other words, for x = a, 



d n y 
dx*' 



dx daf 

must all have the same values, respectively, taken from the equations 
of both curves : and 



dx" 



must have different values. 



172. When the Order of Contact is Even, the Curves cross at the Point 
of Contact ; but when the Order is Odd, they do not cross. Let us dis- 
tinguish the ordinates of the two curves by 

T=€f»(x), and y = if/(x). 

In the figures Y refers to the full curve, and y to the clotted curve. 

If Y—y has the same sign on both sides of P, as in the first 
figure, the curves do not cross at P; but if Y—y is positive on one 
side of P and negative on the other, the curves do cross at P. 

Let OM=a, MM 1 = h. 

Then AQi= Y-y==4>(a + h) - if/(a + K). 









F 


> 


p ^ 


^^ 


_ 








or 


^y^ 










rs 










/ 








/ / 


Qs 








' / 










/ 















R^rT 








. y%~ 


. F 


s 




// 






// 






// 






QyY 






/ 








/ 








// 


p 2 






7 









J M 9 M Mj X 

Expanding by Taylors Theorem, 



M. M Mi 



a \o 



^(a)-^'(a)-|V"(a)-|V"(«) 



13 



(1) 



208 DIFFERENTIAL CALCULUS 

Suppose the contact of the first order ; then 

<£ (a) = ^ (a), <j>'(a) — ^'(a) , and (1) becomes 



^=| 



*»(a)-f'(a) 



+ |T*'"(«)-f "(«)] + •- • (2) 



For sufficiently small values of h the sign of the lowest power de- 
termines that of the second member, and hence the sign of P Y Q X will 
remain unchanged when — li is substituted for h, giving P 2 Q 2 , as in 
the first figure. 

Thus when the contact is of the first order, the curves do not cross 
at the point of contact. 

Again, suppose the contact of the second order ; then 



™=i 



cf>"(a) = «//"(a), and (2) becomes 

V "'(«) - *"'(«)] + jj [+"(«) - r («)] + • ■ 



Now PiQi will change sign with J\, so that P 2 Q 2 and P X Q X will have 
different signs, as in the second figure. 

Thus when the contact is of the second order, the curves cross at 
the point of contact. 

By similar reasoning the general proposition is established. 

It may be of service to the student, in connection with this prin- 
ciple, to think of two curves as having two consecutive common 
points, when they have contact of the first order; as having three 
consecutive common points, when they have contact of the second 
order ; as having n + 1 consecutive common points, when they have 
contact of the nth order. 

An odd number of common points implies the crossing of the 
curves, but where there is an even number of common points, the 
curves do not cross. 

173. Osculating Curves. Contact of the nth order requires that y 
and its first n derivatives should, for some definite value of x, have 
the same values for both curves. 

This implies n + 1 conditions. 



ORDER OF COXTACT. OSCULATING CIRCLE 209 

The equation of the straight line, y = ax + b, having only two 
arbitrary constants, can satisfy only two of these conditions. Hence 
a straight line can have contact of the first order with a given curve, 
and cannot, in general, have contact of a higher order. 

The equation of the circle x 2 + y 2 -\- ax + by + c = 0, having three 
arbitrary constants, can satisfy three of the conditions. Hence the 
circle may have contact of the second order with a given curve. 
Such a circle is called the osculating circle. 

Similarly, the parabola, whose equation contains four constants, 
may have contact of the third order ; and the general conic, whose 
equation contains five constants, may have contact of the fourth 
order with a given curve. These are called the osculating parabola 
and the osculating conic. 

174. Order of Contact at Exceptional Points. Although the tangent 
has generally contact of the first order, it may at exceptional points 
of a curve have a contact of a higher order. 

For example, since the tangent at a point of inflexion crosses the 
curve, it follows from Art 172, that the order of contact must be 
even. Hence at a point of inflexion the tangent has contact of at 
least the second order. 

The osculating circle, which has generally contact of the second 
order, has a higher order of contact at points of maximum or mini- 
mum curvature, as, for example, the vertices of an ellipse. It is 
evident from the symmetry of the ellipse with reference to its ver- 
tices, that no circle tangent at these points would cross the curve at 
the point of contact. Hence, by Art. 172, the order of contact is 
odd, — at least the third. 

175. To Find the Coordinates of the Centre, and Radius, of the Oscu- 
lating Circle at Any Point of a Given Curve. 

Let the equation of the given curve be 

y =/(»)• 

The general equation of a circle with centre (a, b) and radius r, is 
(x-a?+(ij-by-=^ (1) 



210 DIFFERENTIAL CALCULUS 

Differentiating twice successively, we have 



HW +(y - b)d A=°- 



(2) 
(3) 



From (3), y - b = 



dx 2 



(4) 



From (2), 



x— a = 



dy 

dx 



1 + f^V" 
1 dx 



c^y 
dx 2 



(5) 



Substituting (4) and (5) in (1), 



1 



iff 

dx 



dx 2 ) 



(6) 



Hence 



dy 
dx 



a — x — 



x\_ \dx 



d*y 
dx 2 



, . \dx) fTs 

> h = y+ ^f-' () 
dx 2 



and 



Wt 



(8) 



C£# 2 



ORDER OF CONTACT. OSCULATING CIRCLE 211 

In these expressions, x, y, — , — , refer to (1), the equation of the 

circle ; but since the osculating circle bv definition has contact of the 
second order with the given curve, these quantities will have the 
same values if derived from the equation of this curve y=f(x), and 
applied to the point of contact. 

By comparing (7) and (8) with the expressions for a, ft, and p, in 
Arts. 163, 165, it is evident that the osculating circle is the same as 
the circle of curvature. 

176. At a Point of Maximum or Minimum Curvature, the Osculating 
Circle has Contact of the Third Order. 

If we regard equation (8) in the preceding article as referring to 
the given curve y =f(x), we have as a condition for a maximum or 
minimum value of r, 

^ = 0. 



dx 



We thus obtain from (8), 
dx\da?) 



M2' 



dx* 



3 dyfdy 

from which tlL = * x[ ' ] f\ . 

dx* im(<&\* 
\dx) 



Again, if we regard (8) as referring to the osculating circle 
(x-af+(y-by=r\ 

we shall also have — = 0, 

dx 

since r is constant for all points on the circle. 



212 DIFFERENTIAL CALCULUS 

Thus we obtain, both for the curve and the circle, the same ex- 
pression (1) for J, and since -^ and — \ in the second member of 
dx 3 dx dx 2 

(1) have, at the point of contact, the same values for both curves, it 

d 3 v 
follows that — t has likewise the same value. Hence the contact is 
dx 3 

of the third order. 

EXAMPLES 
1. Find the order of contact of the two curves, 
y = x?, and y = 3 x 2 — 3 x + 1. 

By combining the two equations, the point x = 1, y = 1, is found 
to be common to both curves. 

Differentiating the two given equations, 

y = x 3 , y = 3 x 2 — 3 x + 1, 

dx ' 

dx>- bX 



^ = 6, 
da? ' 



dx 


= 6aj- 


-3, 


d 2 y_ 
dx 2 


= 6, 




d 3 y_ 
dx?~ 


= 0. 





dy 



When x = 1, ^ = 3, in both curves ; 

dx 

when x = 1 , ^4 = 6, in both curves ; 

dx 2 

but —4 has different values in the two curves. 
dx 3 ^ 

Hence the contact is of the second order. 

2. Find the order of contact- of the parabola, 4?/= a? 2 , and the 
straight line, y = x — 1. Ans. First order. 



ORDER OF CONTACT. OSCULATING CIRCLE 213 

3. Find the order of contact of 

9y = a*-3x* + 27 ) and 9// + 3.r = 2S. 

Ans. Second order. 

4. Find the order of contact of the curves 

y = log(x — 1), and x 2 - 6a; + 2y + 8 = 0, 
at the common point (2, 0). ^l»s. Second order. 

5. Find the order of contact of the parabola, 4,y = x~ — 4, and the 
circle, or -f y- — 2 y = 3. Ans. Third order. 

6.- What must be the value of a, in order that the parabola, 
y = x + l + a(x— l) 2 , 

may have contact of the second order with the hyperbola, 

xy — 3x — 1? Ans. a—— 1. 

7. Find the order of contact of the parabola, 

(x-2a) 2 +(y-2ay=2xy,- 

and the hyperbola, xy = a 2 . ^4?is. Third order. 



CHAPTER XIX 
ENVELOPES 

177. Series of Curves. When, in the equation of a curve, different 
values are assigned to one of its constants, the resulting equations 
represent a series of curves, differing in position, but all of the same 
kind or family. 

For example, if we give different values to a in the equation of 
the parabola y 2 = £ ax, we obtain a series of parabolas, all having a 
common vertex and axis, but different focal distances. 

Again, take the equation of the circle (x — a) 2 -f (y — b) 2 = c 2 . By- 
giving different values to a, we have a series of equal circles whose 
centres are on the line y = b. 

The quantity a which remains constant for any one curve of the 
series, but varies as we pass from one curve to another, is called the 
parameter of the series. 

Sometimes two parameters are supposed to vary simultaneously, 
so as to satisfy a given relation between them. 

Thus, in the equation of the circle (x — a) 2 + (y — b) 2 = c 2 , we may 
suppose a and b to vary, subject to the condition, 

tf + b 2 = tf 

We then have a series of equal circles, whose centres are on 
another circle described about the origin with radius fc. 

178. Definition of Envelope. The intersection of any two curves 
of a series will approach a certain limit, as the two curves approach 
coincidence. Now, if we suppose the parameter to vary by infinitesi- 
mal increments, the locus of the ultimate intersections of consecutive 
curves is called the envelope of the series. 

214 



ENVELOPES 



215 



179. The Envelope of a Series of Curves is Tangent to Every Curve of 
the Series. 

P Q 




Suppose L. My N to be auy three curves of the series. P is the 
intersection of M with the preceding curve L, and Q its intersection 
with the following curve N. 

As the curves approach coincidence, P and Q will ultimately be 
two consecutive points of the envelope and of the curve M. Hence 
the envelope touches M. 

Similarly, it may be shown that the envelope touches any other 
curve of the series. 



180. To find the Equation cf the Envelope of a Given Series of Curves. 

Before considering the general problem let us take the following 
special example. 

Required the envelope of the series 
of straight lines represented by 

. m 

y = ax -f — 

a 

a being the variable parameter. 

Let the equations of any two of 
these lines be 



y = ax + 



m 



and y = (a + 



a + h 



(1) 

(2) 



From (1) and (2) as simultaneous 
equations, we can find the inters 
tion of the two lines. Subtracting (1) from (2), 




216 DIFFERENTIAL CALCULUS 

= to- hm , 

a(a + A) 

or = £ ™ — (3) 

From (3) and (1), we have 

x= m , (2a + h)m . . . . . (4 ) 

which are the coordinates of the intersection. 

Now if we suppose h to approach zero in (4), we have for the ulti- 
mate intersection of consecutive lines 

m 2 m 

ar a 

By eliminating a between these equations we have 

y 2 = 4 ma?, 

which, being independent of a, is the equation of the locus of the in- 
tersection of any two consecutive lines, that is, the equation of the 
required envelope. 

The figure shows the straight lines, and the envelope, which is a 
parabola. 

181. We will now give the general solution. 
Let the given equation be 

f(x,y,a) = 0, 

which, by varying the parameter a, represents the series of curves. 

To find the intersection of any two curves of the series, we com- 
bine 

f(x,y,a) = Q, (1) 

and f{x, y,a + h) = (2) 



ENVELOPES 217 

From (1) and (2), we have 

f{x, ?/, a + h)-f(x, y, a) _ n ^x 

A ' (; 

and it is evident that the intersection may be found by combining 
(1) and (3), instead of (1) and (2). 

When the two curves approach coincidence, h approaches zero, 
and we have, by Art. 15, for the limit of equation (3), 

|-/(.r,2/,«) = (4) 

Thus equations (1) and (4) determine the intersection of two con- 
secutive curves. By eliminating a between (1) and (4) we shall 
obtain the equation of the locus of these ultimate intersections, 
which is the equation of the envelope. 

182. Applying this method to the preceding example, 

. m 
y = ax + — , 
a 

we differentiate with reference to a, and obtain for (4) Art. 181, 

= x --- 
a 2 

Eliminating a between these equations gives the equation of the 
envelope, 

y 2 = 4: r mx, as found'in Art. 180. 

183. The Evolute of a Given Curve is the Envelope of its Normals. 
This is indicated by the figure of Art. 166, and the proposition 

may be proved by the method of Art 181, as follows : 

The general equation of the normal at the point (V, y') is by 

Art. 148, x -x! + &( S -y l ) = 0, (1) 



218 DIFFERENTIAL CALCULUS 

dn' 
in which the variable parameter is x', the quantities y', -~, being 

dx' 

functions of x'. Differentiating (1) with reference to x', we have 

From (1) and (2) we find for the intersection of consecutive 
normals, 

, , \dx' 

y=y + 



x = x' — 



dy' 
dx' 



dx' 2 

HSf\ 



d 2 y' 
dx' 2 



As these expressions are identical with the coordinates of the 
centre of curvature in Art. 165, it follows that the envelope of the 
normals coincides with the evolute. 

EXAMPLES 

1. Find the envelope of the series of straight lines represented by 
y — 2 mx -f- m 4 , m being the variable parameter. 

Differentiating the given equation with reference to m, 

= 2^ + 4m 3 . 
Eliminating m between the two equations, we have for the envelope, 
16?/ 3 4-27x 4 = 0. 

2. Find the envelope of the series of parabolas 

y 2 = a (x — a), a being the variable parameter. Ans. ky 2 = x 2 . 

3. Find the envelope of a series of circles whose centres are on 
the axis of X, and radii proportional to (m times) their distance 
from the origin. Ans. y 2 = m 2 (x 2 + y 2 ). 



ENVELOPES 219 

4. Find the evolute of the parabola y 2 = 4ax according to Art. 
l^o. taking the equation of the normal in the form 

y = ni (x — 2 a) — a m s . Ans. 21aif = 4 (x — 2 a) 3 . 

5. Find the evolute of the ellipse — + ^=1, taking the equation 
of the normal in the form 

by = ax tan <j> — (a 2 — b 2 ) sin <j>, 

where <£ is the eccentric angle. 

Ans. (ax)* + (by)* = (a 2 — b 2 ) \ 

6. Find the envelope of the straight lines represented by 

x cos 3 + y sin 3 = a(cos 2 0)*, 

6 being the variable parameter. 

Ans. (x 2 + y 2 ) 2 = a 2 (x 2 — y 2 ), the lemniscate. 

7. Find the envelope of the series of ellipses, whose axes coincide 
and whose area is constant. 

The equation of the ellipses is 

£,+£=i a) 

a- cr 

a and b being variable parameters, subject to the condition 

ab = k 2 , (2) 

calling the constant area -n-k 2 . 

Substituting in (1) the value of b from (2), 

K + &-1, (3) 

a- k 4 

in which a is the only variable parameter. Differentiating (3) with 
reference to a, we have 

-^+^ = (4) 

a 8 Jr 

Eliminating a between (3) and (4), we have 
4 xhf = k\ 



220 DIFFERENTIAL CALCULUS 

Second Solution. Differentiate (1), regarding both a and b as 

variable. 

x?da j y 2 db = ^ / 5 n 

a 3 b 3 



Differentiating (2) also, we have 

b da + a db = . (6) 



From (5) and (6), we have 



9 9 

£=£ (7) 

a 2 b 2 W 



From (7) and (1), 



t = t = \ (8) 

a 2 b 2 2 W 



Substituting (8) in (2), 

4 x 2 y 2 = & 4 . 

8. Find the envelope of the circles whose diameters are the double 
ordinates of the parabola y 2 = 4 ax. Ans. y 2 = 4 a (a + x). 



9. Find the envelope of the straight lines - + ^ = 1, 

a b 

when a n + b n = k n . ■ _ JL _ _^ _n_ 



X 2 V 2 

10. Find the envelope of the ellipses — \-¥- — l 9 

a 2 b- 

2 2 2 

when a + b = fc. -4ns. # 3 + 2/ 3 = & 3 . 

11. Find the envelope of the circles passing through the origin, 
whose centres are on the parabola y 2 = 4 ax. 

Ans. (x + 2a)y 2 + x? = 0. 



ENVELOPES 221 

12. Find the envelope of circles described on the central radii of 
an ellipse as diameters, the equation of the ellipse being 

a~ o- 

13. Pind the envelope of the ellipses whose axes coincide, and 
such that the distance between the extremities of the major and 
minor axes is constant and equal to k. 

Ans. A square whose sides are (x ± y) 2 = k 2 . 



INTEGRAL CALCULUS 



:>**c 



CHAPTER XX 
INTEGRATION. STANDARD FORMS 

184. Definition of Integration. The operation inverse to differ- 
entiation is called integration. By differentiation Ave find the dif- 
ferential of a given function, and by integration we find the function 
corresponding to a given differential. This function is called the 
integral of the differential. 

For instance; 
since 2xdx is the differential of x 2 , 

therefore x 2 is the integral of 2xdx. 

The symbol I is used to denote the integral of the expression 
following it. 

Thus the foregoing relations would be written, 



d{x 2 ) = 2xdx, p2xdx = x\ 



It is evidently the same thing, whether we consider this integral 
as the function whose differential is 2xdx, or the function whose 
derivative is 2x. 

As regards notation, however, it is customary to write 



| 2zdx = :c 2 , and not j 2x = xr. 



223 



224 INTEGRAL CALCULUS 

In other words, 

/d 
is the inverse of d, and not of — 
' dx 

Thus the general definition of I <1>(x)dx is that function whose 

differential is <^{x)dx ; the symbol | denoting " the function whose 

differential is," in the same way that the inverse symbol, tan -1 , 
denotes "the angle whose tangent is." 

Integration is not like differentiation a direct operation, but con- 
sists in recognizing the given expression as the differential of a 
known function, or in reducing it to a form where such recognition 
is possible. 

185. Elementary Principles. 

(a) It is evident that we may write 

I 2 x dx = x* + 2, or I 2 x dx = x 2 — 5, 

as well as J 2 x dx = x 2 ; 

since the differential of x 2 -f- 2, as well as of x 2 — 5 is 2 x dx. 

In general J 2 x dx = x 2 + (7, 

where C denotes an arbitrary constant called the constant of integra- 
tion. 

Every integral in its most general form includes this term, 
+ 0. 

(6) Since d(u ±v ± w) = du ± dv ± div, 

it follows that 

I (du ± dv ± dw) = I du ± i dv ± I div. 



INTEGRATION. STANDARD FORMS 225 

That is, we integrate a polynomial by integrating the separate 
terms, ami retaining the signs. 

(c) Since d(au) = adu, 

it follows that | aclu = a j dw. 

That is, a constant factor may be transferred from one side of the 
symbol J to the other, without affecting the integral. 

186. Fundamental Integrals. Since integration is the inverse of 
differentiation, to integrate any given function we must reduce it to 
one or more of the differentials of the elementary functions, ex- 
pressed by the fundamental formulas of the Differential Calculus. 
Corresponding to these formulae we may write a list of integrals, 
which may be regarded as fundamental, and to which all integrals 
should, if possible, be ultimately reduced. We shall then consider 
in this chapter such examples as are integrable by these formulae, 
either directly, or after some simple transformation. 

I. frcfe-Jf-L. 

J n-f-1 

II. Jy = logu. 

III. Ca*du= ,— . 

J log a 

[ »• /• 

r . j cos u 



e"du = e u . 



V. I cos u du = sin u. 



VI. I sin u du — — cos u. 



226 INTEGRAL CALCULUS 

VII. ) sec 2 u die = tan u. 

VIII. I cosec 2 w du = — cot w. 

IX. | sec u tan udu = sec w. 

X. J cosec u cot icdu = — cosec w. 

XI. j tan it du = log sec u. 

XII. I cot udu = log sin u. 

XIII. j sec u du = log (sec w + tan w) = log tan f - + 

XIV. I cosec u du = log (cosec u — cot w) = log tan - 

<S ^t-xt C du 1 , i u 1 ,_i u 

XV. j = -tan -1 -, or = cot 1 -- 

J vr + a 2 a a a a 

, S vvt C du 1 i u — a 1 , a 

\X XVI. -^ - = — -log-— — , or = —log- 

J vr — a? 2 a u-\- a 2 a a 

XVII. f 

XVIII. r_J?^= log (u+v^tf) 
•/ VV ± a 2 y 



7T , tfc 

2 



+ u 



XVII. I — = = sin * — , or = — cos l - 

Vcr — u 2 a a 

du_ 

± 



ATT AT C WU J- _1 '<*. 

XIX. I —-sec -, or = cosec 

J M V w 2 _ a 2 a a a a 



XX. f r/ " = 

•^ V2 aw — w 2 



= vers ' - 
a 



INTEGRATION. STANDARD FORMS 227 

INTEGRALS BY I. AND II. 
187. Proof of I. and II. 

To derive I., 
since d(u n+l ) = (?i -f- 1) u n du, 

therefore 

« n+1 = f (n + l)u n du = (n + 1) Cu n du, by (c), Art. 185. 

/ u n+l 
u n du = • 
n + 1 

Formula II. follows directly from / 

,j i du 
d log u = 

u 

It is to be noticed that I. applies to all values of n except n = —1. 
For this value it gives 



u° 

v l du = — =oo 



/• „ 

Formula II. provides for this failing case of I. 

EXAMPLES 

Integrate the following expressions : 



1. 



CaHkx. 



If we apply L, calling u = x, and n=4; then du —dx. Then 
we have 

j A <hr = '— + C, adding the constant of integration C, according to 
(a), Art. 185. 



228 INTEGRAL CALCULUS 

2. C(x 2 + l)hdx. 

If we apply I., calling u = x 2 -j- 1, and n = - ; then cfa*=2 #d#. 

We must then introduce a factor 2 before the icda?, and conse- 
quently its reciprocal - on the left of | 

f (<e* + l)*a> da; = i f (» 2 + 1)*2 x dx, by (c), Art. 185. 

_l (a; 2 + l) f _0g_+l}j 
2 3 3 

2 

3 r (a; 2 --a 2 )cfa = l / %3a 2 - 3a 2 )(fo 
' J x*-3a 2 x 3 J a; 3 -3a 2 x 

= |log (a 3 - 3 aV) = log (or 5 - 3 aV)* + C. 
o 

By introducing the factor 3, we make the numerator the differ- 
ential of the denominator, and then apply II. 

4. f(2x»-3x«+12x 3 -3)dx = — - ?jl + 3x*-3x + C. 

%s O 4 

5. f(»f_l- + 2_2V 3»t_ 3 ^_ 1 _ 2 _ 
c/ \ ^f ct- 5 a;y 5 2x* 

6. f(aj 2 -2)Vcto=- } -^- 8 + — -2* 4 + <7. 
*/ 10 4 o 

7. f(a^-2) 3 a?da; = ^~ 2 ) 4 -|-0. 

«y 8 



INTEGRATION. STANDARD FORMS 229 

CI \ }\ 3 7 9ctiv$ , 9oU x- , ~ 



2a 



10 (7 *F 4 /« 3 Y. 4a# 12 a£ , , 11,4a*', „ 
v * a ^ 13a* 7a* 5^ 

11. jy»i + aJ -lY(to = ^ + i^+ 6» + 12**-3ar* + 0. 

12. fcil S ^ = ^^ + 32 ,-log y+ a 

J 7/ 3 2 

J x i 7 4 

14. JV + 1) 5 x 2 da; = ( a?3 + 1 ) 6 + C. 

15. j%.x~ + 6)Wi\ 16. C(ax -\-b) n dx. 

17. C(aa? + b)*a?dx. , 18. f (oaf + ft^a?"- 1 da?. 

21. Ju-i-l/"^'. 22. J(a + logOy- 



230 
23. f. 



INTEGRAL CALCULUS 



e x -\-x e 



-dx. 



24. j (a — log 10 x) 



dx 
x 



25 - $\J^zf + *£- 



/■ 



e 2x da. 26. ( sin 5 cos (9 ^6>. 



27. f (e 2 * + sin 20)(e 2 + cos 20)d0. 



28. I tan 6 x sec 2 x dx. 29. I sec 5 a tan x dx. 

30. f (sin m -f cos w 0) sin cos 6 dO. 

31. f(sec + tan 0) 10 sec <9d0. 

32. ("(sin <£> + cos <£) w (sin 2 <f> - cos 2 <£)d<£. 33. C(a x + &) 3 a x da. 



34 



r 



sin -1 a £fa 



35 



' J(l + 



da 



x~) tan -1 a 



A rational fraction, whose denominator is of the first degree, may 
be integrated directly or after being reduced to a mixed quantity. 



36 - /iSri log(4 *- 3 > +c - 



37 



• Jfffi^ = ^-^ 1( 



2 4 



33 



J 2x-l 3 2 2 4 ■ V y 



INTEGRATION. STANDARD FORMS 



231 



on Cost -f b , ax . 6 s — a 2 1 n , \ , r\ 

39. ; dte = — + , log (&as + a) + C. 

J bx + a b b~ 



40 C^LtJ^ civ = - 4- — + « 2 -« + 1 a 8 log (» - a) + G. 
J x — a 3 2 

41 . f (- r + flV <& = f + 2 as - fcc + (a - &)' 9 log (x + 6) + <7. 

«y .17 -}- — 



42. f ( A ' + f ^ s <*b = - + 2 oas 2 + 7 a 2 * + 8 a 3 log (x - a) + C. 
J x — a 3 



INTEGRALS BY III. AND IV. 

188. Proof of III. and IV. These are evidently obtained directly 
from the corresponding formulae of differentiation. 



EXAMPLES 



. f(^ + a * + 3b-*ydx=^.+ i^ -l^-f 



5 log a 2 log b 



2. f <>« + e- ax fdx = -\ e -^ + 3e ax -3 e~ ax - 

*J Cl\ o . 



a\_ 3 



+ 0. 



&e 4x 7a 



, r I j , , 1 /2a 2 9a 3 , 18a 6 



fa 



4. I L cto = — 6 e 3 h C 

J %&* 4 2 



232 INTEGRAL CALCULUS 

5. C(e x+a - e ax+h ) dx. 6. C(e 5inx cosx — a cos2x sm2x)dx. 

i 

7. C(e x \ + -\dx. 8. fte tan * sec 6> -e sec nan0) sec 0d<9. 

J e 2 * J J V ; log(«6 2 ) 

log (a m b p ) 

J v ; 21og6 log (a&) 2 log a 

INTEGRALS BY V.— XIV. 

189. Proof of V. — XIV. It is evident that V. — X. are obtained 
directly from the corresponding formulas of differentiation. 
To derive XI. and XII., 

tan udu = — | = — log cos u — log sec u. 

J cos u 

/, , (* cos udu i 
cot udu= I — : = log sin u. 
J sinw 

To derive XIII. and XIV., 

/' -j _ r sec u (tan a 4- sec u) du _ /~ sec u tan u du + sec-u du 

J sec u + tan u J sec it 4- tan-% 

= log (sec u 4- tan u) . 

/«„„«„ - ,7 Tcosec w (— cot z* 4- cosec w) aw 
cosec udu= I i l — 
J cosec u— cotu 

= log (cosec u — cot w). 



INTEGRATION. STANDARD FORMS 233 

By Trigonometry, 

2 sirr - 
1 — cos u 2 . u 

cosec u — cot u = — : = = tan - • 

sm a o • u it 2, 

2 sin -cos- 

— _ 

If we substitute in this, ^ + u for u, 
we have sec u + tan w = tan ( - + - 

Thns we obtain the second forms of XIII. and XIV. 

, Cf • o - sb\ 7 cos 3 aj . sin 5 x 

1. I sm 3 x + cos o x — sin - ax — — - -\ — • 

+ 2cos^+C. 

sm — -*- 1- cos — !— ■ cto = — m cos ^ + n sm — ^— + C. 

J \ m n J m n 

o CI 4- sin mx 7 1 ,, , x , AT 

3. | — : — dx = — (tan mx -\- sec mx) 4- C. 

J COS' 0KC ??i 

4. I (sec 5 x — tan 5 a?) sec 5 x dx. 5. I (sec 2 4- tan 2 6) dO. 

J \cos-6 smdj J 

c> f* vers y 

8. I — ^— — da; = cos x — 2 log (1 4- cos #) 4- C. 

J sm./; y 

9. C^pdx. 10. '.fl?™!*** 
•/ c J sin-x 

n r seo»rf» = 1 , ,,, tan 4> + b) + a 
J a sin £ + & cos <£ a 



234 INTEGRAL CALCULUS 

12. I (tan x — cot# + l) 2 dsc = tan a; + cot cc — 3x — 2 log sin 2 x + C. 

13. I (sec 2x-\- tan 2 a? — cot 2 a?) 2 c?# = tan 2x -\- sec 2x — - cot 2 a; 

— log tan ce — 4 x -j- (7. 

14. I (sec <f> + cosec <f> — I) 2 dcf> = 4> + tan <£ — cot <£ 

+ 2Iog* + c ° s f+a 

1 + sin <£ 

The following may be integrated after trigonometric transforma- 
tion. 

15. Csin*xdx = *-™^+a 



4 



\. I sin 2 a? 

1 a C 2 7 » , sin 2 x . ^ 

lb. | cos^ a; cfe = - -\ — ■ 1- C. 

J 2 4 

17. . ( vers 2 a; dx. 18. j sin 2 x cos 2 x dx = — — + C. 

J J x 16 

in f. a . n -, n sin (m — n)0 sin(m + %)0 , n 

19. I sin mO sin nO dO = - ) '- — ^ — — J — + C. 

J 2 (m — n) 2 (m -f w) 

nr\ C a a 7/1 ■ sin(m — ?i)0 , sin(m + w)d , ^ 

20. | cos m0 cos w0 c?0 = - — \ J — + -ttt — r~ + "• 

J 2{m — n) 2(m-j-w) 



oi C - a a ja • cos (m — w)0 cos (m + w)0 ' n 

21. j sm m0 cos w0 d0 = '- —-) ! — f- -f C\ 

J 2(m— w) 2 (m+n) 



on T K o j sin 3 x . sin 1 x , n 

22. | cos t> a? cos 2 £(&«:= — 1 — \- C. 

J 6 14 



INTEGRATION. STANDARD FORMS 235 

23. f sin (3 x + 2) cos (4 • + 3) dx = < **(* + *) - C os( ^+^ + C. 

nA T ■ .-> o 7 cos 6 a; cos 4 a; cos 2 sc , ^ 

24. I sm .r sin 2 x sm 3 .r dx = — — -f C. 

J 24 10 8 

25. C^MdO. 26. rg?°i- fl rfg = 2si n g + 2gin3fl + C. 
J sin J sin o 

27. f — ^ — = tan0-sec0 + O. 28. f-^-. 

J 1 + sm J vers 

-dx — — cosec a? —cot x — log vers a; + C. 

vers x 

C cos a; cfce „ , 

VI + sm x dx = I — — 2 VI — sin a; + C. 

J V 1 — sin .x* 

31. fVveraada?. 32. f- ^ = J- log tan (^ + ^+ C. 

J J sin + cos V2 V2 87 

INTEGRALS BY XV. — XX. 
190. Proof of XV. — XX. 
To derive XV., 



29 



30 



c_±!_ = i r^^ = i f_W = i taa -i« 

J it 2 + a 2 a .7 i i ?L a 1 (-] a a 

cr \aj 



To derive XVII., 

du 



I •= I — SID-1-. 

•J Va 2 — <•- ^ ■}_'_!! a 

\ a 2 



236 INTEGRAL CALCULUS 

To derive XIX., K " 

du 



/du 1 r a 1 ,u 
3;:;: I _ — SeC _ ' 
wVw 2 -a 2 uJ u ly?_ 1 a a 
a^a 2 



To derive XX., 

du 



/du C a 

^/2au — u 2 J \u u 2 

\2 2 

* a or 



vers" 



Since tan- 1 ^ = 7r -cot- 1 ^, 

a 2 a 



it is evident that d tan -1 - = d ( — cot -1 - ) • 



Hence either expression may be used as the integral in XV. 

In the same way we obtain the second forms of XVII. and XIX. 

The formulae XVI. and XVIII. are inserted in the list of integrals, 
because their forms are similar to XV. and XVII., respectively, with 
different signs. 

To derive XVI., 



u 2 — a 2 2 a\u — a u + aj 

hence 

/ du 1 Cf du du \ 

u 2 — a 2 2aJ \u — a u + a) 



h [log (w - a) ~ log (M + a)] = £ log hi 



INTEGRATION. STANDARD FORMS 237 

Or we may integrate thus : 



r (hi _ \_ r r -du _ (7?/ 

J ir~ a 2 2 a J \a — u a + u 



= A [log (« - u) - log (a + «)] = A log a 



j a a + U 



To derive XVIII. 



assume -\vF±a- = z, a new variable. 

Then u 2 ±a 2 = z 2 , 

2udu = 2zdz; 

du dz du + dz 



therefore 



z u u + z 



Hence f^ ( = C d " + (fe = log (u + z) ; 

1 1S ' J v^±^' = log (w + v%2 ± a2) - 



EXAM PLES 



J4.r + 9 6 3 J4a; 2 -9 12 °2a; + 3 

3. I — ====== = -sm * \- C. 

J VT-25^ 5 2 

4. P dj; = I log (5 a; + V25 ar - 4) + C. 

5. f r? ' r = — log (ars/5 + Voa?+l) + 0. 

•^ V5 a^ + 1 Vo 



238 INTEGRAL CALCULUS 

J3-12x 2 12 & 2a?-l^ 



/ dy g C dw q C dx 

12 f + 3* 'Jm<; 2 -3° ' J 3^-5* 

^ V3* 2 -2 ' J V2 - 3 a 2 . J V3~a^T2 



13. C dx = lsec-^+G 

•^ ^V9ar — 4 2 2 



» /: 



dec -l 2 a? ' ^ 

= vers x h C. 



V ma? — a? 2 m 

15. f ^ =-sec- 1 g + g 

J&VaV-16 4 2 



16. f ^ = ± vers" 1 



dx 1 _, -8 x 

_ ====r = - vers l — 

V7 a? — 4 a? 2 2 7 



17. P da; = sin" 1 log Vx + C. 

J a?V4— (log a?) 2 

18 -/SS^ = i l0g( ^ +9) ^i tatl "1 +a 

■ J4a? 2 -5 8 4V5 2x+V5 



20. f 3 a; ~ 2 da? = - 3 V9 - x 2 -2sm- l ^+C. 
J V9-a? 2 3 



a 



INTEGRATION. STANDARD FORMS 239 

21. f x _± ___ c?.r = ^ ^T4 -f 3 log (x + V.^+4) + (7. 

J V .r + 4 



22. C 5 x 1 dx = 5 V3 X s - 9 - — log (a V3 + V3 ^ - 9) + C. 

J V3.r-9 3 V3 



23. 



J cr siu- + 6- cos- «6 6 



24. f rf » = -Ltaii-'gg4 + C. 
J 1 + cos- <#> y 2 V2 



25. r si ° g ^ =CO gV«*JUft 

•/ A/Sp.ns 2 fl4-a.siTi 2 fl V 2 / 



V3 cos- + 4 sin- 



26. f / S 2£J_^1 - = -^Iog(3cos2s+V9cos 2 2a;-4) + (7. 

*^ Vo cos 1 ' 2 .1- — 4 sin- 2 i " 

27. f e _!l±^! tfa. = 1 log ( e 2x + x + J_ tan -i _£_ + c 

The same formulae may be applied to integrals involving 
- ax + /j or — x 2 + ax + b, by completing the square with, the 
terms containing x. Thus, 

Ja* + 6a?+13 J (x + 3/ + 4 2 2 

29. f ** = f _ «** ^sin-'g^ + C. 
J v/8 + 4 a; — 4 .r J a 9 - (2 jc - l) 2 2 3 

30. f r7x = Jl. W (3 a? - 2 + V9ar>- 12 a; + 6) + C. 

-'a a7+ 2 V3 



240 INTEGRAL CALCULUS 



31. f *5 = -4,tan-i^4?+C. 

J 2 ar — 3 a? + 5 A /si V31 

32. 



2 x 2 - 3 a; + 5 V31 V31 

/' dx 
— , when a = 4 ; when a = 6 ; when a = 8. 
or — ax + 9 

33. f- -*? = 1 tan-^^±^±J + a 

J (a> + a) 2 + (a; + 6) 2 Va 2 + & 2 Vet 2 + b' 1 

O/i r ^ 1 • -1 4 JB — 1 . n 

34. I — === ^ = = - sm 1 — + C. 

35.- f eos2ecW = ±log sin2e +a 

J sin 2 20+msin20 2m & m + sin2 



36. f dx =sm-i 2x - a - b +C. 

J V(a-a)(&- a-) ■ a-& 



37. f- *L = 2 tan -i(^ + a + &)V3 +a 

J (a? + a) 3 — (a? + 6V 3 a /q („ _ m* a — b 



(x + af-(x + by V3(a-6) 2 



CHAPTER XXI 



SIMPLE APPLICATIONS OF INTEGRATION. 

TEGRATION 



CONSTANT OF IN- 



Before continuing the integration of functions, we will consider 
the relation of integration to the determination of the area bounded 
by a given curve, and show how the constant of integration may be 
determined. 

191. Derivative of an Area. Let y=f(x) be the equation of a given 
curve OP v Suppose a point to move along the curve starting from 
P ( „ and let x, y, be the coordi- 
nates of any position P. 

At the same time the ordi- 
nate of the moving point 
starts from the position 
P^lf,,, and sweeps over or 
generates a certain area. 
When the point has moved 
to P, this area is PqMqMP. 
Denote this area by A. 

A is a function of x, and it 
will now be proved that its derivative with respect to x is equal to y. 

Give to x the increment A.? = MN. 




Then 



AA = PMNQ. 

AA > y Ax, and AA < (y + Ay) Ax, 



-r->y> 

Ax 



AA 

and ■— < y + Ay. 

Ax 



Hence 



rl_A 

dx 



= Lim 



AA 



.'/• 



In case the curve descends from P to Q, the above inequalities 
will be reversed, but the result will be the same. 

241 



242 INTEGRAL CALCULUS 

192. Area of Curve. Let it be required to find the area PqMqMxP^ 
between the curve, the axis of X, and the two ordinates P M Q and 

Pal- 
let 0M Q = a, and 0M X = b. 

dA 

From the preceding article — = y. 

dx 



Hence A = j ydx= i f(x)dx. 

Let • Cf(x)dx=F(x) 



0} 

then A = F(x) + G (1) 

To determine C, we have the condition that A begins when x = a ; 
.that is, A — when x = a. 

Hence = F(a) + C, C=- F(a). 

Substituting in (1), A = F(x) - F (a) = P M MP (2) 

It is to be noticed that C is determined by the initial value a of x, 
corresponding to the initial ordinate PoM Q . 
If now we let x = b in (2), we have 

A = F(b) - F(a) = PqMoM^ 

For example, let the given curve be the parabola y 2 = x. 

Then A= Cydx= f x idx= — + G. ... (3) 

To determine G, A = when x = a. 

o 3 



SIMPLE APPLICATIONS OF INTEGRATION 248 

Substituting in (3), 

A=Z^-^ = P<>M Q MR (4) 

To find PJUfP.. let s= b in (4). 

3 3 " 



EXAM PLES 

1. In the curve of Ex. 1, p. 24, show that P OJIP, = ~' 

Also P 1 M 1 M 2 P 2 = ^- 

2. Find the area included between the equilateral hyperbola 
2xy = cr. the axis of X. and two ordinates x = a, x = 2 a. 

Ans. cr los: V2. 



3. Find the area included between the witch of Agnesi (Art. 126), 
the axes of X and I", and the ordinate x = 2 a. Ans. ircr. 

4. Find the area included between the catenary (Art. 128), 
the axis of X, and the ordinates x = a, x = 2a. 

2 

A n s. — (ft 2 — e + e _1 — e~ 2 ) . 

5. Find the area of one arch of y = sin x. Ans. 2. 

6. Find the area included between the parabola x* + y- = a* 
(Art. 129), and the axes of X and Y. Ans. 



<r 



7. Find the area included between the semicubical parabola 
". - = • A rt. 130 • the axis of 1", and two abscissas, y = 8 a, gl =* 27 a. 

A 633 



244 



INTEGRAL CALCULUS 



Conversely, instead of determining the area from the integral, we 
may find the integral from the area, when it can be obtained geo- 
metrically from the figure. For example : 

8. Find | Va~ — x 2 dx, by means of the curve y = Va 2 — x 2 , circle 
about 0, radius a. 

A = BOMP = OMP+ BOP 




= -xy-\ <}>= -x-y a~ — xr -\ sin - • 

2 y 2 * 2 2 a 

If the initial ordinate, instead of OB, 
had been some other ordinate, we should 
have had 



A = - Va 2 - x 2 + - sin- 1 - + C, where C is independent of x. 
Hence A= i ydx— j Va 2 — sc 2 cZas = *| Va 2 — a,* 2 + ^- sin -1 - + (7. 

9. Find j (3 a? + 2) eta, by means of the line y = 3x + 2. 

10. Find J V2 ax — x 2 dx, by means of the curve y = V2 ax — x' 2 . 

Ans. CV2ax-x* dx= X -^ V2aa-a 2 -f £ sin" 1 ^^ + C. 
J 2 2 a 

193. Other Illustrations. In order to further illustrate the deter- 
mination of the constant of integration, we will work three examples, 
involving geometrical or physical properties. 

Ex. 1. Determine the equation of a curve through the point (4, 3), 
at every point of which the slope of the tangent is equal to the recip- 
rocal of twice the ordinate of the point of contact. 



SIMPLE APPLICATIONS OF INTEGRATION 



245 



By the hypothesis 


dy_ 1 
dx 2y 


from which 


2ydy = dx. 


Integrating, 


y- = x + a 



(1) 



2 = £-8 




This equation represents a series of parabolas whose axes coincide 
with the axis of x. 

If now we impose the additional condition that the curve must 
pass through the point (4, 3), its coordinates must satisfy equation 
fe), giving 9 = 4 + C, 0=5. 

The equation of the particular curve is therefore 

y 2 = x -h 5. ' 



Ex. 2. A body starting from rest, with a given initial velocity ^ , , 
moves with a constant acceleration g. Find the space passed over 
in any time. 

dv 



In Art. 19, acceleration 



dt 



Here 
Integrating, 



rj } dv==gdt. 



dt 
v = gt+C. 



From the conditions of the example, v = v when t = ; therefore 
v = 0+C, C=Vq. 



246 INTEGRAL CALCULUS 

Hence v = gt-\- v . 

ds 
Since v = — (Art. 18), ds = gtdt-\- % dt. 

tit 

Integrating, s = - gt 2 + v$ + C. 

From the conditions of the example, s = when t = ; therefore 
O=0, and s = - #£ 2 + v £ is the complete solution. 

Ex. 3. A body is projected at an angle a with the horizon, and 
with a velocity v . Eind the equation of its path. 

Represent the horizontal and vertical components of the velocity 
by v x and v y respectively. Then, since gravity is the only force act- 
ing on the body, we have 





^ = 0, and 

dt 


dt * 


Integrating, 


«L=C, 


v y = -gt+0. 


When t = 0, 


v x = v cos a, 


v y = v sin a. 


Hence 


v x = v cos a, 


v y = -gt + v l) sina-, 


at is, 


dx 

-r- = i< cos a, 
dt 


-^- = — gt + v sin a. 

(*5 



Integrating, x = v t cos a + C, 2/ = — - gtf 2 + v £ sin a + C". 

When t — 0, x and ?/, and therefore C and C, are zero. 
Hence x = v £ cos a, and y = — -gf-\- v t sin a. 

Eliminating £ between these equations, we have as the equation of 
the path of the projectile, 

qxr 



?/=#tan a 



2 v? cos 2 a 



This evidently represents a parabola whose axis is parallel to the 
axis of Y. 



SIMPLE APPLICATIONS OF INTEGRATION 247 



EXAMPLES 

4. Find the equation of the curve whose subnormal (Art. 146) 
has the constant value 4, and which passes through the point (1, 4). 

Ans. y 2 = 8x + 8. 

5. Find the equation of the curve whose subtangent (Art. 146) 
is twice the abscissa of the point of contact, and which passes through 
the point (2, 1). Ans , x = 2y 2 . 

6. The slope of the tangent to a curve at any point is , and 

the curve passes through the point (3, 2). Find its equation. 

Ans. 4ar + 9?/ 2 = 72. 

7. Find the equation of the curve whose polar subtangent (Art. 
153) is 3 times the length of the corresponding radius vector, and 

which passes through the point (2, 0). Ans. r = 2e s . 



8. Find the equation of the curve whose polar subnormal (Art. 
153) is 3 times the length of the corresponding radius vector, and 
which passes through the point (2, 0). Ans. r = 2 e w . 

9. Find the equation of a curve through the point (3, — j, in which 

the angle between the radius vector and the tangent is half the 
vectorial angle. Ans. r = 6(1 - cos 6). 

10. A balloon is ascending with a velocity of 20 miles an hour. 
A stone dropped from the balloon reaches the ground in 6 seconds. 
Find the height of the balloon when the stone is dropped, 

Ans. 400 ft. 

11. If a particle moves so that its velocities parallel to the axes 
of X and Y are ky and. kx respectively, prove that its path is an 
equilateral hyperbola. 



248 INTEGRAL CALCULUS 

12. A body starts from the origin of coordinates, and in t seconds 
its velocity parallel to the axis of X is 6 1, and its velocity parallel 
to the axis of Y is 3 1 2 — 3. Find (a) the distances traversed parallel 
to each axis in t seconds ; (b) the distance traversed along the path ; 
(c) the equation of the path. 

Ans. (a) x = 3f; tj = f-3t. 

(b) s = t 3 + 3 t. 

(c) 27tf = xfr-$y. 

13. If a body, projected from the top of a tower at an angle of 
45° above the horizontal plane, falls in 5 seconds at a distance from 
the bottom of the tower equal to its height ; find the height of the 
tower (gi = 32). Ans. 200 ft. 

14. When the brakes are put on a train, its velocity suffers a con- 
stant retardation. If the brakes will bring to a dead stop in 2 min- 
utes a certain train running 30 miles an hour, how far from a station 
should the brakes be applied, if the train is to stop at the station ? 

Ans. Half a mile. 



CHAPTER XXII 
INTEGRATION OF RATIONAL FRACTIONS 

194. Formulae for Integration of Rational Functions. On examin- 
ing the fundamental integrals in Art. 186, it will be seen that only 
four apply to the integration of rational algebraic functions, I., II., 
XV.. and XVI. ; and of these only I., II., and XV. are independent, 
since XVI. depends directly upon II. 

It will be shown in this chapter that by these three formulae any 
rational function can be integrated. The integration of a rational 
polynomial has been explained in Chapter XX. We will now con- 
sider the integration of rational fractions. 

195. Preliminary Operation. If the degree of the numerator is 
equal to, or greater than, that of the denominator, the fraction 
should be reduced to a mixed quantity, by dividing the numerator 
by the denominator. 

For example, 

a*-2o* = 1 2.t 2 + 1 
x' + l a^ + l ' 



2a 5 -3tf 4 + l Q o . _2ar 3 + 3a 2 + l 

— - — — z x — o -\ • 

' X* + X~ X* + XT 

The degree of the numerator of the new fraction will be less than 
that of the denominator. 

The entire part of the mixed quantity is readily integrable, and 
thus the integration of any rational fraction is made to depend upon 
the integration of one whose numerator is of a lower degree than the 
denominator. 

240 



250 INTEGRAL CALCULUS 

196. Partial Fractions. A rational fraction is integrated by de- 
composing it into partial fractions, whose denominators are the 
factors of the original denominator. The complete discussion of 
Partial Fractions belongs to Algebra. We shall only consider here 
the form of these partial fractions and the processes of determining 
them. 

Factors of the Denominator. It is shown by the Theory of Equa- 
tions that a polynomial of the nth. degree, with respect to x, may be 
resolved into n factors of the first degree, 

(x — a x ) (x — a 2 ) (x — a 3 ) • ••(x — a n ). 

These factors are real or imaginary, but the imaginary factors 
occur in pairs, of the form 

x — a + &V— 1, and x — a — b V — 1, 

whose product is (x — a) 2 + b 2 , a real factor of the second degree. 

It follows that any polynomial may be resolved into real factors 
of the first or second degree, and only such factors will be considered 
in the denominators of fractions. 

There are four cases to be considered. 

First. Where the denominator contains factors of the first degree 
only, each of which occurs but once. 

Second. Where the denominator contains factors of the first 
degree only, some of which are repeated. 

Third. Where the denominator contains factors of the second 
degree, each of which occurs but once. 

Fourth. Where the denominator contains factors of the second 
degree, some of which are repeated. 

197. Case I. Factors of the Denominator all of the First Degree, 
and none repeated. 

The given fraction may be decomposed into partial fractions, as 
shown by the following example, 



/ 



x 2 + 6 x — 8 , m 
X s — 4 x 



INTEGRATION OF RATIONAL FRACTIONS 251 

sume 

.r + .r - 8 = x 2 + 6 .v - 8 AB C § ; ^ 

where ^1. 5. C are unknown constants. 

Clearing (1) of fractions, 
x* + 6x-8 = Ax(x + 2)+Bx(x-2)+C(x-2)(x + 2) ... (2) 

=(A + B+C)x i + 2(A-B)x-4:a 

Equating the coefficients of like powers of x in the two mem- 
bers of the equation, according to the method of Undetermined 
Coefficients, we have 

A + B+C=l, 2(A-B) = 6, _4C=-8. 

whence 4 = 1, B = -2, C=2. 

aj* + 6a>-8 1 2,2 

Hence — — = (--? 

x 3 — 4:X x — 2 x-i-2 x 



and 



r x* + 6x-S dx = 1()g (flj ^ 2) _ 2 log (flJ + 2) + 2 log 



= log^^- 2 ). 
8 (a + 2)* 

The following is a shorter method of finding A, B, C: 
Suppose the denominator of the given fraction to contain the fac- 
tor x — a, not repeated. Then the fraction may be expressed as 

fix) = A | tfx) 
(x — a) <f> (a?) x — a <f>(x) 

Hence &* = A + (x - a) ^M . 

This being an identical equation is true for'all values of x. 



252 INTEGRAL CALCULUS 

If we put x = a, we have A = ^f^, since by hypothesis <£ (x) does 

not vanish when x = a. ^ w 

' Thus we have the following rule : 

To find A, the numerator of the partial fraction , put x = a 

x — a 

in the given fraction, omitting the factor x — a itself. 

For example, having written equation (1), we find A by substitut- 
ing x = 2 in the given fraction — l —^ , omitting the factor 

x-2. This gives (x-2)(x + 2)x 

4 + 12-8^ 

4(2) 

To find B, substitute x = — 2, omitting the factor x-\-2. 

4-12-8 = _ 2 
-4(-2) 

To find C, substitute x = 0, omitting the factor x. 



EXAMPLES 

The constant of integration C will be omitted in the examples in 
this chapter, and the following chapters on the integration of func-' 
tions. 

J x- — 3#-j-2 3 2 #—1 

2 f (x*+x+l)dx = l 1 (x + l)(x-3r 
*Ja?-4a? + a; + 6 12 * (o;-2) 28 

3 r(zo+iyd W = i loy , (2^ + i)(2^-i)» 



h 



INTEGRATION OF RATIONAL FRACTIONS 253 



l)(.i-+3)(.c + 5) 8 ° (.c + l)(x + 5) 



_1, (8a»-l)»(8«»-8) 



" J (2as-l)(3ir-l)(3*-.2) 18 (2.i--l) a 

6 /♦ as + ta (fa= _^L^lo g (to _ „) +— *_,log (ax-6). 

J (CKB — 0)(OSE — CI) GO— 0- CIO — (T 

y r Cv + aYclr = 1 lQ C2 x + <Q (s- «y 

J (x+d)(x + b) b-a oV ' a-b &v ; 



9 r «y = i log fr+io'fr-i)- 1 



J (oV — ft 2 ) (6\r — a 2 ) 2 a6 ° (ax — b)(bx -j- a) 

., r fa + i)dx _ l (,- + 5)%,---y 

' J (a? - 19/ - i (x + 8f 360 ° (a; + 3) 8 (a; - 1) 5 " 

12. C >^m:L_ 

J1j;>-17a- ! + 4.»; 



= jLlog^| -1 :^:,:.<.,- + l)(2a.-l)'] + ilog*. 



251 INTEGRAL CALCULUS 

198. Case II. Factors of the denominator all of the first degree, and 
some repeated. 

Here the method of decomposition of Case I. requires modifica- 
tion. Suppose, for example, we have 

a 3 + l 7 
ax. 



J x(x 



(x-iy 

If we follow the method of the preceding case, we should write 

a? + l = A B C D 

x(x — I) 3 X x — 1 x — 1 x — 1 

But since the common denominator of the fractions in the second 
member of this equation is x(x — 1), their sum cannot be equal to the 
given fraction with the denominator x(x — l) 3 . To meet this objec- 
tion, we assume 

x? + l _A , B G D 



x(x-Xf x (x-iy (x-iy x-1 

Clearing of fractions, 
X s + 1 = A(x -l) 3 + Bx + Cx(x - 1) + Dx(x - 1) 2 

= (A + D)x* + (- 3 ^ + <7- 2 D)x 2 + (3 A+ B-C + D)x-A. 

Hence A + D = l, (1) 

= SA+C-2D = 0, (2) 

3A + B-C+D = 0, 

-4=1. 

Whence- A= -1, 23 = 2, (7=1, D = 2. 

Therefore (f + *) = - - + 



x(x - 1) 8 x (x-iy (x - 1) 2 x-1 



INTEGRATION OF RATIONAL FRACTIONS 255 

Hence f f + * 8 dte== -logs- 1 i_ + 21«g>(»--l) 

J jc(a; — 1)° (.x- — 1)- x — 1 

.r . , (x-1) 2 
= - ; ~rr- 2 + log* *-■ 

The numerators A and 5 may be determined by the short method 
given for Case I., and then C and D may be found by (1) and (2). 



EXAMPLES 

»' ar — ar 2 2x'- x a; 



aftfc; 2 - 3 a; .It a? - 1 



2 r a*dx = 2-3a; 1 lQ 

J (a + l)(aJ - l) 3 4(a? - 1)- 8 ° a; + 1 



3. f *^=_1 + h * 

J x\ ar — 4 )-' ar — 4 4 



_ 4)2 x 2 -44 ° ar 9 - 4 

r (19 x-S2)dx 1 3 j „ 2 a; - 3 

J (4;/- + l;(2a;-3) 2 4(2 a; -3) 4 ° g 4.i- + l 

f ^ g = * +J_l 0g ^2. 

J(9ar-4/-' 18(9 ar- 4) 216 8 3 x + 2 



-fa , a 2 — 3 aas . t a? 2 



— ".':/ a; — a 

7. f ' - J '* = , + W—**l - -J* + 8 log (, _ 1). 



256 INTEGRAL CALCULUS 

8 



. C(*±*\*<te = x - i a ~ 6 >' - 3 ( a ~ ^ + 3(a - b) log (x + b). 
J\x+b) 2(x + b) 2 x + b K J &\ t j 

9 C xdx = 2x ~ 1 + — W x-2 + V3 

" J(aj»_4aj + 1)» 6(a 2 _4x + l) 6 V3 \_2-W 

10 f (* + a + &) 3 ±_(_V_ + J*\ 

J (a+a) 2 (x + 6) 2 " (a-6) 2 V« + a x + b) 



. 3 a& 2 — b 3 -, / . N . a 3 — 3 a 2 6 t , , , N 

199. Case III. Denominator containing Factors of the Second 
Degree, but none repeated. 

The form of decomposition will appear from the following 
example, 

f- 



'5cc + 12 , 
■dx. 



x(x 2 + 4) 

w 5 x + 12 A . Bx+C /-, N 

We assume ^ = - + , y , (1) 

a?(ar + 4) a; a;- + 4 

and in general for every partial fraction in this case, whose denomi- 
nator is of the second degree, we must assume a numerator of the 
form Bx + O. 

Clearing (1) of fractions, 

5 x + 12 = (A + B)x 2 +Cx + 4,A. 

A + B = 0, 0=5, 4 A = 12. 

Whence .4 = 3, B = -3, 0=5; 

5 a; + 12 3,-3^+5 



therefore 



x(x 2 + 4) 03 x 2 + 4 



f-3«+s fe= _ 3 r^. +6 r 



a 2 + 4 J x' + i Ja? + 4 

-|log(^ + 4) + ?tan-^. 



INTEGRATION OF RATIONAL FRACTIONS 257 

Hence f ° '*'•/" Vn dx = 3 log . * + r, tan_1 % ' 

Take for another example, 

r (2x 2 -3x-3)dx 

J (x-l)(.r--2.i' + 5)' 

This fraction is decomposed as follows : 

2a?-3x-3 1 3.v-2 



(a — l)(jr — 2# + 5) a? — 1 x 2 — 2x + 5 

r *3x-2)dx = r (3x-3)dx C dx 
J g*-2x + 5 J x*-2x + 5 J x*-2x + 5 



= |log(^-2 a; + 5) + |tan- 1 ^i 



r <2*>-3x-S)dx =1 (*-2s+5)t + l tan -i£ = l, 
J (.-, _ l) (^ _ 2 2 + 5) ° a? - 1 ^2 2 

The integration of any fraction with a quadratic denominator like 

the preceding. ( \ ' ~~ — V— , may be shown as follows : 
* °' J x*-2x + 5' J 

Having written the denominator in the form (a; + a ) 2 + b 2 , we have 

r ( px + g)dx __ C p(x + a)dx . r (g — pa)dx 
. (x + o.f + b 2 ~J {x + ay + W J (x + a) 2 + b 2 



= £ log [(x + a)^ + &*] + *=» tan- 2±5 . 
J b b 



258 



INTEGRAL CALCULUS 



r32_ 
" J 4 ar 



EXAMPLES 



32 x> + 3 7 8 a; 3 A ( 1, 



+ 3a? 



4^ + 3 



+ 3V3tan- 



V3 



/» (2s»-aQ(to ^l l0 . (^-l)(^ + 2) + 2V2 tan - 1 J 

' J x 4 +ar 9 -2 6 & 



(®+iy 



V2 



■x : 



• '4^-30 + 1, _6£-l 
2rf + rf * t ""2^" + 



2 a? 



- + 3V2tan- 1 (a-V2). 



„ r rfdx x 3 , 1 i x-1,1, _i 
4 ' J^I=3 + I l0g ^Tl + 2 tan *• 



When the given fraction and the denominators of the partial 
fractions contain only even powers of x, they may be regarded as 
functions of x 2 , and we may assume A, B, C, etc., as the numerators 
of the partial fraction. 

In the following example, the partial fractions may be assumed as 



B 



x- + a 2 x 2 + b- 



5 C x * dx 

J (x 2 + a 2 )(x 2 



ft 3 tan -1 a 3 tan 

(per + cr) (x- + &-) or — b~ V 



= x + 



s 



(1 - x 2 )do 



i^l^-i 3x 



= - 1 tan -1 2 x — tan -1 - ] = - tan" 
(4a> 2 + l)(a? + 4) 6 V 2 J 6 2 + 2 a,- 2 



Zr> 



J (a¥+5 2 )(5¥ + a 2 ) ' a6 a&(l - a 2 ) 

J a;(ar 9 - 6 a? + 13) 26 ° 8 jc 8 13 



, _i x - 3 



INTEGRATION OF RATIONAL FRACTIONS 259 



9 r (3x>-2x-20)dx = l lo/ , (2^-6a? + 6y 



+ — tan- 1 ^- - 2 tan- 1 (2 a? - 3). 
V3 V3 2 ^ ; 



10. lAJlna ly- 1 / _u_l,t. a ,-i2y±l 



I f * = - losj -A3 A- -j tan 

Jt?-1 6 V + y.+ * V3 V3 

11. f^±g^ = li og ^ + a ? + l + vg tail - 1 W3. 
J s* + 3^-4-1 4 & .r- - x + 1 2 1 - a? 

12. I — -— = : log ^— H -tan ] . 

J ^ 4 + l 4V2 iv 2 + w^2+l 2V2 I-™ 2 

1Q r dx 1 ,1^ ay-1 2 , _!2aj-l 

13. |- — - = — ; r-+Tlog tan 1 

Jtf-^ + ^-l 6(* + l) 4 *x+l 3 V '3 V3 



200. Case IV. Denominator containing Factors of the Second De- 
gree, some of which are repeated. 

This case is related to Case III., as Case II. to Case I., and requires 
a similar modification of the partial fractions. 

For illustration take 



s 



2*+*+*^ 



O-s + l) 2 

We assume 

2s 3 + s 2 + 3 _ A x + B Cx + D 
(a^-f-l) 3 (a?+l)* x* + 1 

* + 3 = Oc 8 + A* + M + C)a> + B + Z>. 

-4*-2, /* = 2, C=2, £==1. 



260 INTEGRAL CALCULUS 

Therefore lt±_t±l = -2* + 2 + l*±A 

(a^ + 1) 2 (a; 2 +l) 2 ^^ 2 + l 

/» -2a + 2 /» 2a?da? 2 f dx 

J (x' + iy J (^ + i) 2_r v (o^ + i) 2 

a: 2 + l "*" J <V+1) 2 ' 

To integrate the last fraction, we use the following formula of 
reduction, 

J CISC J- U/ . /(y q\ i CltC &. 

(x 2 +« 2 ) w ~ 2 (n - 1) a 2 [_(o5 8 + a 2 )' 1 " 1 ^ * ' } J (a 2 + a 2 )-^ ' 

/die 
— — by making it 
(ar-f-cr)' 1 

depend upon j — tt~~~^ — ' -By successive applications the given 

integral is made to depend ultimately upon | _ which is 

-! J x~ -\-d 2 

itan- 1 -. 
a a 



* This formula may be derived as follows : 

[ * 1 = A[x(x 2 + a 2 )-"] = (x 2 +a 2 )" w - 2 nx 2 (a? + a 2 )-"- 1 

|_(x 2 +a 2 ) M J dx 



j^ r £ d 

dx _, 

= (x 2 + a 2 )-" - 2 to [(x 2 + a 2 ) - a 2 ] (x 2 + a*)-"- 1 
= (1 - 2 to)(x 2 + a 2 )~ n + 2 na 2 (x 2 + a 2 )-"" 1 . 



Integrating both members after multiplying by dx, 

* = (l-2n)f (?X +2na 2 f , **, . 

(x 2 + a 2 )" v y J (x 2 + a 2 ) n J (x 2 + a 2 ) n + l 

2 %a 2 f *5 = x_^ , 2 n _ x) f_dx § 

J(x 2 +a 2 ) w +i (x 2 +a 2 ) w v y J (x 2 +a 2 ) M 

Substituting for to, n — 1, we have 

2 (w - 1) a 2 f ^ = ? + (2 to - 3) (—-J**- — . • 

v J J (x 2 -f a 2 )" (x 2 + a 2 ) "- 1 v J (x 2 + a 2 )"- 1 



INTEGRATION OF RATIONAL FRACTIONS 
Substituting in the formula n = 2 and a 2 = 1, we have 

J ur + 1)- 2 [_.r + 1 J s 2 + 1 
C — ° r 4- 9 1 



2G1 



■ h - tan -1 sc ; 

2(.r + l)^2 



A partial fraction of the form ^ x + ^ , by substituting 

|> + a) 2 + 6 2 ]»' y 

4. a = z, becomes P "; ^ ^ 9a , the integration of which has already 



been explained. 



C^ + ft 2 }' 



For example, if x-3 = z, j £> =J __*, 



5 + 16 f 

4<V 2 4-3V 2 J i 



f?2 



(^ + 3y 



By the formula of reduction, 
J (* 2 + 3) 3 12L(z 2 + 3) 2 + J (* 2 + 3) 2 J 

12(^ 2 + 3) 2 4 6[_z 2 + 3 Jz 2 + 3j 



1 . _i z 
H — - tan * 



+ 



12(z 2 4-3) 2 24(z 2 4-3) 24 V3 V3 



/5z 
(z 2 



5z + 16 7 162-15 , 2z 

dz = tt— - — — , + 



tan 



-i 2 



(2 2 4-' 12(z 2 + 3) 2 ' 3(z 2 4-3) " 3V3 

(*« + 2)tfa 160,-63 



e f- 
J (a 



6a?4-12) 8 12 (a 2 — 6a?4-12) 3 

2 (a? - 3) 2 tan -i a?-3 

+ 3(s*-6a + 12) 3V3 Un V3 



262 INTEGRAL CALCULUS 

EXAMPLES 



4^ + 3 ,_4 t x* + 5x~-2^_ 1 .,2* 



d% = "' ,.„ — ^r^ H = tan" 



(4a; 2 + 3) 3 8(4a; 2 +3) 2 16 V3 V3 



For the following example, see note preceding Ex. 5, Case III. 

4 r 36s»(s? + l) a + 25a,* x* + x 

' J (4a 2 + 9) 2 (9a; 2 + 4) 2 * 2(4a?+9)(9aj 8 + 4) 

156 6 — 6 x 2 






r ^ + Sx-21 , 3(q; + 7) 

J (^ + 4x + 9) 2 ' 2(3 2 +4«4 



+ 9) 

^ ^ V5 



i C ( xS - 2 ) clx = a^ + 4 

J ( a ^ + a; _|_i)( a j2 + a ,_ h 2) 2 7(ar* + a; + 2) 



+ ^ tan- 2£±1 _ _2_ ^ 2x±t 

7V7 V7 V3 V3 






(a^ + l) 2 a; 3 + l 2 °a; 2 -a; + l V3 

da; = 12a; 2 + 36a; + 29 

[( X + 2) 4 - (a? + l) 4 ] 2 ~" 2 (2 a; + 3) (2 a 2 + 6 x + 5) 

Stan" 1 (2 a; + 3). 



CHAPTER XXIII 
INTEGRATION OF IRRATIONAL FUNCTIONS 

201. We have shown in the preceding chapter that the integral 
of any rational function can be expressed in terms of algebraic, loga- 
rithmic, and inverse-trigonometric functions. 

We shall now consider the integration of irrational functions. 

202. Integration by Rationalization. Some integrals involving 
radicals may be integrated, by reducing them to rational integrals 
by a change of variable. This is possible, however, in only a very 
limited number of cases. This process is sometimes called integra- 
tion by rationalization. 

p 

203. Integrals containing (ax + b) q . Such an integral may be 

rationalized by the substitution ax -{-b = z q . 



For example, take ( — - 



x 2 clx 



(2x + sy 



Assume 2 x + 3 = z i , x= — - — , dx = — — 



Then 



/ :rax 
(1 «. _i_ < 



(2x 



+ 3)* J z SJ 



=l(t-¥ +9s )=¥(7-¥ +9 )=if2 f2a;+3)S(8a;2 - 18a:+81 > 



203 



264 INTEGRAL CALCULUS 

Another example is i ,J _ ' • 

J Va? + i 

Assume x = z 2 , dx = 2zdz. 

Then C_^x_ = f2^ = 2 C,U^ z + 1 _J_\ dz 

J Vx + i J » + i J \ 2 + v 

= 2|~- - - + ^ - log (2 + 1)1 = ^ - x + 2 a* - log (a>* + 1) 2 . 

204. Integrals containing (ax + 6)% (aaj + &)%••• . In this case 
the integral is rationalized by the substitution ax + b = z n , where 
n is the least common multiple of q, s, •■•, the denominators of the 
fractional exponents. 

Take, for example, j 



' (x - 2)* + (x - 2f 



Assume x — 2 = z 6 , dx = 6z 5 dz, 

( x _ 2 )i = z 3 , (x-2)i = z\ 

= 6p- 2 -2-}-log(z + l) = 3(aj-2)*-6(aj-2)* + 61og[(a;-2)* + l]. 



EXAMPLES 



L f ^ + L dx = 2Vx-2 + V2tan- 1 J^=J- 
•^ xVx — 2 * 2 



INTEGRATION OF IRRATIONAL FUNCTIONS 265 

4. (V* + * l +-1 civ = ^ + 2 ^ - 4 .^ + 4 log (s* + 1). 

5 f !?cbf = 4?/ 3 -6?/ 2 -6?/-l 
' J (4y + l)* 24(4y + l)* 

6 f ^- = 2 + 2 log (V2w-1 + 1). 

J W + V 2 80 — 1 A 2 M> — 1 + 1 

8. f.r Vrt7+6 da; = 2 ( q - Y + 6 ) a (15 a *».« _ 12 a&a . + g &2\ 
J 105 a 3 

9. f * = 2(3a; + l)*-4tan-^+i)L 

J (3* + l)* + 4(3a; + l)* 2 

10. f A -' + X o ~ 1 dte = 2 Va^fl - log (as + 3) - 2 V2 tan" 1 J^±^. 

11. f fa ~ 2 ) fe = g(2s-3)Mlog (2 *- 3)l + 3 

^ (2z-3)* + 6a;-9 4 8 (2 a -3)* 

4 V3 



12 



. r 

J V'2:/: + l+Vo;-l 
= 2V2^Tl - 2V^1 + 2 V3 (tan-iyjxEl _ tan" 1 J?^±3 

= 2V2^+1 - 2V^1 + Vsfcos- 1 ^^ - oc»-il=*Y 

^ x + 2 x + 2J 



266 INTEGRAL CALCULUS 

13. 



f ;— ^-5 =2xMlog(2**-l)+!log(** + 2) 

J (2aj*-l)(a>*+2) 9 9 



16 V2, , «* 

— — - tan -1 — ; = 
9 V2 



J (a? + l)* + l 3 



4 



--log(l + V^+l). 

205. Roots of Polynomials of Higher Degrees. — In the rationaliza- 
tion of irrational integrals we now pass from roots of binomials of 
the first degree to roots of polynomials of higher degrees. 

Here rationalization is limited to the square root of an expression 
of the second degree. 



206. Integrals containing Vtf 2 + ax -f b. This may be rational- 
ized by the substitution 



For example, consider I 



■yx 2 + ax + b = z — x. 
dx 



x -y/x 2 — x + 2 
If, following the method of the preceding articles, we assume 



■^x 2 -x + 2 = z, x?-x + 2 = z 2 , 

the expression for x, and consequently that for dx, in terms of z, 
will involve radicals. This difficulty is avoided by assuming 



■Vx 2 — x-\- 2 = z — x, — x-\-2 = z 2 — 2zx, 
cancelling x 2 in both members. 



z 2 -2 



dx = 2(z 2 -z + 2)dz ^ 



2z-l' (2z-iy 

l -z + . 
2z-l 



V^^x~T~2 = z-x = z2 



INTEGRATION OF IRRATIONAL FUNCTIONS 267 

Hence, 

2(z 2 -z + 2)c7z 
dx (2z-l) 2 r<2dz 1 , z-V2 



J «-*/rf_*j.9 s*-2 y-2 + 2 Jz 2 -2~A>° g 





(2* 


-I) 2 




2 2 - 


_ o 


Z 2 — 2 


+ 2 


2z 


-1 


2z- 


-1 



. rw -_ x +2 I z--^ , z -z + ^ Jz--Z V2 z + V2 
<-/ 2z-l 2z-l 

Substituting z = Va,- 2 — x + 2 -f- a?, 

c& 1 , Var 2 - x + 2 -f x - V2 






x-Vx 2 — x + 2 V2 Var' — a; + 2 + a? + V2 



207. Integrals containing V — x 2 -f a# + b. This may be rational- 
ized by the substitution 



V—x 2 + ax+ b = V(« — a;) (/? + ») = (« — #)z or = (/? + #)z, 

where « — x and /? + a; are the factors of —z? + ax-{- b. 

These factors will be real, unless V— ar* + a# -f- 6 is imaginary for 
all values of x. 

dx 



Take, for example, ( — - 

J a;V2 + 



x — x 2 



Assume V2 + x — x 2 = V(2 — x) (1 + x) = (2 — aj)z. 

l + z = (2-a0z 2 , a ? = 2 f-" 1 , dx= 6zdz 



z 2 + l' (z 2 + l) 2 

3z 



V2 + a? - a; 2 = (2 - z)z = j^ • 

Therefore, 

r da; = r 2dz _ 1 log gV2-l 

J W2 + X-.X 2 J2z 2 -1 V2 zV2-fl 



2 — a; 
dx 1 , V2 + 2a;-V2-a; 



f ** = -J-log 

^ a^ + aj-ar 2 V2 



^/2 + x-x 2 V2 V2 + 2a; + V2-* 



268 



INTEGRAL CALCULUS 



EXAMPLES 



dx __•, _ x x + Va; 2 + 4 x — 4 



/ dx _ 

a;V'ar + 4 x — 4 



tan 



2. f Va ^ + 4a; cfa = 



a; + V a; 2 + 4 a; 



+ log (a; + 2 + Va; 2 -f 4 a;). 



/ 



da; 



x— a 



. _ T 2\ 2 a 2 V2 asc — a; 2 



(2 aa? — a; 2 ) 



J ^ ; 2(n + l) 2(»-l) 



/; 



— ^±A)^ = _ = 2 ^ tl-x 4 
(3-a;)V3-2a;-a; 2 



tan" 



3 + x V3 

1 2 

rt — cos 

2 yS 3-x 



ij 3-3c 
* 3 + a; 



_ia; + l 2 _i 2 a; 

COS x — r COS * 



/ Va; 2 -f ft 2 
3a? + 4a 



, z 2 + a 2 4 a n , 5o, 3 2! — a 
dx = — — log 2 H log 



62 9 



9 z + 3a' 



where 



= x + Va^ + a 2 . 



208. Integrable Cases. — The preceding articles include those 
forms of irrational integrals that can be rationalized. In general, 
integrals containing fractional powers of polynomials above the first 
degree — except the square root of polynomials of the second degree 
— cannot be rationalized, and cannot be integrated in terms of the 
elementary functions, that is, cannot be expressed in terms of alge- 
braic, exponential, logarithmic, trigonometric, or anti-trigonometric 
functions. 



INTEGRATION OF IRRATIONAL FUNCTIONS 269 

Every integral may be regarded as defining a certain function. 
It has been shown in Art. 192 that if f(x) is any continuous func- 
tion of Xj I f(x)dx is a function of x, which may be geometrically 

represented by an area bounded by the curve y =f(x) ; but this 
cannot always be expressed in terms of the elementary functions. 



CHAPTER XXIV 

TRIGONOMETRIC FORMS READILY INTEGRABLE 

209. It is to be noticed that any power of a trigonometric func- 
tion may be integrated by Formula I., when accompanied by its 
differential. 
Thus, 

/. _ 7 sin w+1 x C » ^ cos n+1 x 
sin" x cos xdx = , I cos' 1 x sin x ax = , 
n + 1 J 7i+l 

/, _ 2 7 tan w+1 x C > n 2 ■, cot w+1 x 
tan w x sec 2 x dx = — . I cot" x eosec 2 x dx = - , 
n + l J Ti + 1 ' 



J sec n 



x sec x tan x dx 



+ 1 J 71 + 

sec w+1 x 

71+1 ' 

cosec M+1 # 



/ 



cosec w a? cosec x cot a; da? = — 



71 + 1 



Having in mind these integrals, the student should readily under- 
stand the transformations in the following articles. 

210. To find J &vn n xdxov I eos n xdx. When n is an odd posi- 
tive integer, we may integrate as in the following examples : 

J sin 5 xdx= I sin 4 x sin x dx = J (1 — cos 2 a,*) 2 sin x dx 

//h o 2 , 4 \ • 7 ,2 COS 3 X COS 5 X 
(1 — I cos^ x + cos 4 x) sin a? ax = — cos a? H — • 
3 o 

270 



TRIGONOMETRIC FORMS READILY INTEGRABLE 271 

Another example is 

| cos G 2 x dx = i cos 2 2 as cos 2 x dx = - J (1 — sin 2 2 x) cos 2 # 2 dx 

1/ • sin 3 2aA 
= - sin 2 x ]• 

A s ) 



211. To find I sin m x cos' 1 x dx. When either m or n is an odd 

positive integer, this form may be integrated in the same manner as 
in the preceding article. For example, 

J sin 4 .r cos 5 x dx = I sin 4 x cos 4 x cos x dx = j sin 4 x (1 — sin 2 x) 2 cos x dx 

/, • 4 o • 6 i • 8 \ 7 sin 5 x 2 sin 7 x , 
(sin 4 x — 2 sm b aj + sm 8 a;) cos x dx == — h 
o 7 



snr a; 



Another example is 
I sin 3 x cos - x dx = I cos T as sin 2 x sin # c7ru = | cos* as (1 — cos 2 #) sin #da; 

J 5 9 



EXAMPLES 

3 cos 5 a; , cos 7 a; 



1 C ■ - i 3 COS 5 25 . 

1 . I sm' x dx = — cos x -f cos J x (- 

J 5 



4 sin 3 a; . 6 sin 5 a? 4 sin 7 a; . sin 9 a; 



o C n 7 4 sin 3 x , 6 sin 5 a? 4 sin 7 a? . 

2. I COS .r r/.v; = sm X h 

J 3 5 7 

o C - -, •'' 7 o as . 4 o«2 5 # 

3. I sin 5 - das = — 2 cos - + - cos 3 - — cos 5 - • 
J 2 23 252 

J T V ^ 9 11 13 15 



272 INTEGRAL CALCULUS 

sin 6 2 sin 8 2 6 



■s 



5. sin 5 2 cos 3 2 <9d<9 = 



12 16 



6 . I (sin 6 x + cos 6 x) sin 3 x cos 2 x dx 



_ cos 3 x 4 cos 5 a? _ 6 cos 7 a; cos 9 x 
~3~ 5 7 "3" ' 



7. I (cos 3 <£ + sin 3 <£) (cos 2 <£ — sin 2 <f>) d<f> 



2 
= sin 2 <£ cos 4> + cos 2 </> sin <£ + - (sin 5 <j> + cos 5 <f>) . 



o rsm 7 ydy cos 6 ?/ 3 cos 4 ?/ . 3 cos 2 y . -, 

8. I £_£ = _ -JL _-*h ^.j-iogsec?/. 

./ cos y o 4 J 

9 C <x**te = fa m + _2 1 . 

J sin 4 x sm a; 3 sin 3 x 



^Q /~ cos 3 a; dx _ _ 2 sin 2 x -f- 6 
•^ Vsin 3 a? 3Vsina? 



11. I (sin™ a; cos 3 a; — cos w a; sin 3 x) dx 

_ sin m+1 x + cos m+1 a? sin™+ 3 x + cos TO+3 a; 
m + 1 m + 3 

12 . I (sin 2 a? + cos 2 a?) cos 3 x dx = - (sin 5 x — cos 5 a;) + sin x cos 2 a;. 



iq /* • , • « 7 4 sin 5 x 8 sin 7 a? 

13. I sin 4 a? sm 3 a; cto = .. 

J 5 7 



TRIGONOMETRIC FORMS READILY IXTEGRABLE 273 

212. To find ( tan" x dx, or ( cot" x dx. 

These forms can be readily integrated when n is any integer. 
J tan" .i- dx = J tan" -2 x (sec 2 x — 1) dx 

= | tan" -2 .i' sec 2 ado;— I tan' 1-2 x dx 



tan x 

tan" -#«#. 



da; 
dx. 



Thus J tan" a; da; is made to depend upon j tan n2 a* cZa;, and ulti- 
mately, by successive reductions, upon J tan a; da; or I dx. 

For example, I tan 5 x dx = J tan 3 x (sec 2 x — 1) 

tan 4 a /*, o 

= ■ I tan-* x 

4 J 

I tan 3 as dx = I tan a; (sec 2 x—l)dx 

tan 2 a; , 
= — log sec x. 

Hence I tan 5 x dx = f- log sec x. 

*s 4 Z 

Another example is 
Ccotfxdx= Ccot 4 x(cosec 2 x-l)dx = -^^- feot 4 x dx 

cot 5 ./* C 9 . , <N , cot 5 a; , cot 3 a; , /• i9 

= I cot 2 a;(cosec 2 aj — l)dx=s \- \- | cot 2 a; 

5 J 5 3 «/ 

cot 5 a; . cot 3 a; . C \ nnat%eAtm 1 , cot 5 a; , cot 8 a; 

= — — 1 h I (cosec- x — 1) dx = - — H cot x - x. 

O o %J Do 



dx 



274 INTEGRAL CALCULUS 

213. To find j sec" x dx or | cosec" x dx. When n is an even 
positive integer, we may integrate as follows : 

j sec 6 xdx= J (teLn 2 x+T) 2 sec 2 xdx= J (tan 4 a? + 2 tan 2 a; + 1) sec 2 a; da; 

tan 5 x . 2 tan 3 x . , 

= — — H htana;. 

o o 

Another example is 

j cosec 4 x dx = J (cot 2 x+1) cosec 2 x dx — — co x — cot x. 

214. To find I tan m x sec n xdx or J cot™ x cosec" x dx. When n is 

an even positive integer, these forms may be integrated in the same 
manner as in the preceding article. For example, 

j tan 6 x sec 4 x dx = I tan 6 x (tan 2 x + 1) sec 2 x dx 
(tan 8 x + tan 6 x) sec 2 x dx = — 1 — . 

When m is an odd positive integer, we may integrate as follows : 
j tan 5 x sec 3 x dx = j tan 4 x sec 2 x sec x tan a? cfcc 

= j (sec 2 x — l) 2 sec 2 x sec aj tan x dx 

= I (sec 6 x—2 sec 4 a; +sec 2 x) sec a; tan a; dx 
sec 7 a; 2 sec 5 x sec 3 a; 



7 5 3 

Another example is 

cot 3 # cosec 5 x dx = ( cot 2 x cosec 4 x cosec a; cot a? dx 



j cot 3 $e cosec 5 x dx = j c 
= j (cosec 6 a? — cosec 4 x) cosec a? cot x dx = 



cosec' a; cosec x 



TRIGONOMETRIC FORMS READILY INTEGRABLE 275 



I tan s x da. 



EXAMPLES 

^_t^ + tan^_ taua; _ f ^ 
i 5 3 



2. Jcotf 5 dx = -| catf | +| cot* | - cot 2 1 - log sin 2 |. 

3. f se c"ya' i/ =^ + 4tau ^ + S^ + 4tan ^ + tan 2 /. 
*/ 9 7 o 3 

4. Jcosec* 3 x dx = - 1 (*£?? + ^1* + 00 V 3 x + cot 3 A 

5. I (sec x — tan x) sec 5 # tan 4 x dx 

1 2 1 

= - (tan 9 x — sec 9 a;) + - (tan 7 #+sec 7 a;) + - (tan 5 x — sec 5 x). 
y 7 o 

6. f(sec 3 <£ + tan 3 <£) 2 cty 

2/. B . . nnn5 ,v . tan 3 <f> 2sec 3 <f> . „. . . 

= - (tan 5 <£ + sec D <£) -\ — -*- -*- + 2 tan <f> — <fi. 

5 3 3 

» /"tan 7 se+l , tan 5 a; tan 4 a; . . . , 

7. I '— dx = — h tan x -f log cos x. 

J tan a- + 1 o 4 

Q rsec 5 ;/* + tan 5 a; 7 , a 2 sec 3 a; , 

o. I cte = tan J £ Hsecar+a;. 

J sec x + tan a; 3 

Q Tsec 6 x + sec 4 a; , tan 2 x . 2 , -, 

9. I -7 dx = — cot 2 a? 4- 3 log tan x. 

J tan 3 a; 2 

10- I 7 .,. #0 = r , f tan- 5 - sin 2 0) -+- log (sin 6 tan 0). 

*/ cosec-<9cot J 2 y 



276 INTEGRAL CALCULUS 

H- J Vsec 3 x tan x (Vsec 5 x— Vtan 5 x) dx 



9 7 y 9 3 3 

= - (tan 2 " ^— sec* x ) -f- ^ (tan* a; + sec" 2 #). 



12. 



J (sec m x* tan 5 x — tan m_1 a? sec 6 x) dx 



_ sec m+4 x — tan OT+4 a; 2(sec m+2 o? + tan m+2 a;) sec m a; — tan TO a; 
m + 4 m + 2 m 

A term I sec m x cosec M x dx may be integrated, when m + n is 

even, by substituting cosec a = -. 

tan x 

13. fsec 5 cosec 3 x dx = *^^ + 3tan ^ - ^^ + 3 log tan x. 
J 4 2 2 5 

14. j (sec 4 a; — cosec 2 x) 2 dx 

tan 7 x , 3 tan 5 a; . tan 3 x , cot 3 x , 

= — ^ 1 — H 3 tan x — — \- cot x. 

too o 



215. To find j sin™ x cos n x dx by Multiple Angles. The integra- 
tion of this form, when either m or n is odd, has been given in Art. 
211. The following method is applicable when m and n are any 
positive integers. 

By trigonometric transformation sin" 1 x cos n x, when m and n are 
positive integers, can be expressed in a series of terms of the first| 
degree, involving sines and cosines of multiples of x. 

If we use the method of Art. 211 for integrating terms with one 
odd exponent occurring during the process, the following formulae 
for the double angle will be sufficient for the transformation of the| 
terms with even exponents : 



TRIGONOMETRIC FORMS READILY 1NTEGRABLE 277 

sm x cos x =- sm A x, 



sin 2 x = - (1 — cos 2 a?), 



cos 2 x = - (1 -f cos 2 #). 



For example, required I sin 4 aj cos 2 s 



2 a efo. 



sin 4 x cos 2 x* = (sin X cos xf sin 2 # = - sin 2 2 .t(1 — cos 2 a?) 

8 



1 - 2 o __ _«_ . 1 



sin 2 2 sb cos 2 x-\ (1 — cos 4 x). 



sin 3 2a; , x sin 4 oj 



tt C • 4 9 7 sm 3 2 a: . 

Hence I sm 4 .i' cos- x ax = h 

«/ 48 



16 64 



EXAMPLES 
n 4 x 



i r • 4 7 1 /3 x • o , si 

1 . I snr x ax = - — sm 2 x-\ — 

2. f cos 4 x dr. = - f^ + sin 2 x + 

3. I sin 2 .'.- cos 2 .'- dx=* - 



sin 4 x 



1 f sin 4 a 

x — 



fsin«aj(ia?= 1 f 7> .,- - 4 sin 2x + sm * 2 ,7 ? + -sin 4 A 
J 16^ 3 4; 



278 DIFFERENTIAL CALCULUS 

5. fcos 6 xdx = — f5ff + 4sin2%- sm32a; +-siii4A 
J 16 \ 3 4 ) 

6. f sin 4 x cos 4 x dx = -i/ 3 a - sin 4 a + !^£\ 

i? /~ 6 -9-, 1 /, , 8 . , . A sin 8 aA 4 

7. I cos b x snr # aa,* == \ox-\- -snr 2 a; — sm 4a; )• 

J ^ 128\ 3 8 / 

8. I sin 8 x dx = — f 4 sin 2 a? 4- - sin 3 2 a + - sin 4 x 

J 16\ 8 3 8 

sin8aA 
+ 64 J' 



CHAPTER XXV 
INTEGRATION BY PARTS. REDUCTION FORMULAE 

216. Integration by Parts. From the differential of a product 
d(uv) = udv + v du, 

we have uv = i u dv + I v du. 

Hence I udv=-uv — Ivdu (1) 

This formula expresses a method of integration, which is called 
integration by x>^rts. 
For example, let us apply it to 









j x log x dx 






Let 


u = 


log a 


j, then dv = 


= xdx; 




whence ' 


du = 


dx 

X 


and v- 


X? 

2* 




Substituting in (1) 


, we 


have 








Jbg; 


x • x dx = log x - — - 
5 2 


-/?• 


dx 

X 








= - loga>- 


X 2 

4* 





• (2) 



279 



280 INTEGRAL CALCULUS 

Integration by parts may be regarded as a process, which begins 
by integrating as if a certain factor were constant. 

Thus in (2), if in | logx • dx we treat log x as if it were a con- 

. x 2 

stant factor, we obtain log x • — • From this we must subtract a 

Z 

new integral formed as indicated by the following connecting lines. 



*x 2 dx 

x 



logx>xdx = logX'--J —■ 

This method of remembering the process may be found useful. 

Another example is j x cos x dx. 

Assuming u = cos x, we have 

I x cos xdx — cos x • — — | — (— sin x dx). 

As the new integral contains a higher power of x than the original 
integral, nothing is gained by this application of the process. 
But if we take u = x, we find 

I x cos x dx = x sin x — I sinscefa; 

= x sin x + cos a?. 

EXAMPLES 



1. J a? 4 logx dx = y flogcc — - 

2. fa (e a * + e~ ax ) dx = - (e ax — e- a *) — i (e ax + e~ a:c ). 
J a a 2 

3. J x (sin 3 x — cos 3x)dx=( (--) sin 3x — (- + -) cos 3 



x. 



INTEGRATION BY PARTS. REDUCTION FORMULAE 281 

4. I x log - + sm - dx = — losr 2 x cos - + 4 sin - • 

J V 2 2) 2 2 4 2 2 



w)- e T"2 ) + 2 



5. fa;(e»»-l) 2 Qte=e te (- 

6 . Clog (ax + 6) dx = fx -f - ) log (oaj + ft) — x. 

7. J( a? + l)log(a ? + 3)cfa ; = a?2 + ^ a; - 3 log( a; + 3)-g + |- 

8 . f sec 4 <£ log sin <£ d<f> = f 1 ^^ + tan ^ log sin <£ - ^L* - 2& 

10> rlog(., + 2) (? ^_log(. + 21 + 1 *±1. 
J (.i' + l) 2 x + 1 *x+2 

11 . ftan- ] - dx = x tan" 1 - - - log (x 2 + a 2 )- 
J a a 2 

12 . pc 2 tan" 1 aj da = - tan" 1 x -- + - log (x 2 + 1). 
c/ 3 6 6 

13. I sin -1 - dx = x sin -1 - + Va 2 — x 2 . 
J a a 

14. f (3 x- 2 - 1) sin- 1 x dx = (x- - x) sin" 1 a; - (* ~ ^ . 



282 INTEGRAL CALCULUS 

n K C • 3 ,1 /cos 3 x \ . sin 3 x . 2 sin x 

15. I a?snrxaa; = x I cosicJH 1 

16. J x (sec 6 x — tan 6 a;) dx = x tan 3 a; -f- — ■ — HL® _j_ i g sec Xt 
„ / 1o S (^ + l) (to = a ,_gl±l l08(e . + 1) . 

18. flog (a + Var 2 + a 2 ) cte = a; log (a + -Vx 2 + a 2 ) 



+ a log (a; -f Va^ 2 + a 2 ) — a?. 

In each of the following examples integration by parts must be 
applied successively. 

19. CaPe-^dx = -^!(4ar 5 + 6a; 2 + 6a; + 3^ 

20. JV - xfdx = £ - ^(4 x - 1) + ^(2x> - 2 x + 1) - ^ 

21 . Car- 1 (log a;) 2 da; = - [(log xf - ?i°S« + -1 ■ 
J n |_ n n 2 J 

22. rx 3 sin2a;^ = ('?^-l N )sin2aj-^--— )cos2x. 

23. Tartan- 1 a;) 2 dx = ^^(tan" 1 x) 2 - a; tan" 1 a; + 1 log (x 2 + 1) . 

x log (a; + a) log (a; — a) dx = - — - — log (aj + a) log (x — a) 



x 2 



Q^!log(x + a)-^-^\og(x-a) + ^ 



INTEGRATION BY PARTS. REDUCTION FORMULAE 283 
217. To find J e™ sin nx dx, and J e ax cos nx dx. 

Integrating by parts, with u = e ax , 

e ax sm nxdx = f-- I e ax cos nxdx. . . (1) 

Integrating the same, with u = sinnic, 

/„ • 7 € ax sin nx n C ax j /o\ 

e ax sin nx dx = e" cos nx dx. . . . (2) 

We see that (1) and (2) are two equations containing the two 
required integrals, j e ax sin nx dx and I e az cos nx dx. Eliminating 
the latter, by multiplying (1) by n 2 , and (2) by a 2 , and adding, gives 

(a 2 + iv) J e M sin nx dx = e™ (a sin nx — w cos nx) ; 

v r ax • -, e^ia sin no; — n cos ?i#) /0 \ 

hence I e"smtia5CW5= — * '- (3) 

J a 2 + n 2 

Substituting this in (1) and transposing, gives 

' a C ax 7 e" (a?i sin ?ia; + a 2 cos na;) 
- I e aj cos nx dx = — * ! z- 

»J a 2 + n 2 

I " C ax 7 e" (w sin as -f a cos wa;) , AS 

hence I e^ cos nx dx = — * — ! L (4) 

J a 2 + ?i 2 w 

EXAMPLES 

The student is advised to apply the process of Art. 217 to Exs. 
1-4. For the remaining examples he may substitute the values of 
a and n in (3) and (4). 



I ' ' sin 5 x cfcr = — (3 sin 5 aj — 5 cos 5 a;), 

/e 3 * cos 5 x dx = — (5 sin 5 as -f 3 cos 5 x). 
34 



284 



INTEGRAL CALCULUS 



2. \ 



/e~ 2x 
e~ 2x sin x dx = — (2 sin x -f- cos a?), 



e~ 2x cos a; da; = — (sin x — 2 cos a?). 



e ax sin aa; dx = — (sin ax — cos ax). 
2a K J 



4. fe hos^dx = ~^ Y2sin? 



*sin 2 a; + cos 2 a;, sin 2 x -f- 5 cos 2 a; 



5. f S — 



13 e s 



(e 2x + sin 2 a?) (e* + cos x)dx — 1 (sin x + 2 cos a;) 



. e* / . o o N 2 cos 3 a; 

H — (sin 2 a? — 2 cos 2 a;) . 

5 3 



7. fe 2 * cos 2 3 x dx = — + — (3 sin 6 a; + cos 6 a?). 



8. I e x sin 2 a; sin 3 a? dx = 



sin a; + cos a; 



5 sin 5 x 4- co s 5 a? 
13 



] 



/ g 2x 
a;e 2x cos xdx = — [5x (sin a; + 2 cos a;) — 4 sin x — 3 cos a;]. 



218. Reduction Formulae for Binomial Algebraic Integrals. These 
are formulae by which the integral, 



j x m (a + bx n ) p dx, 



may be made to depend upon a similar integral, with either m or p 
numerically diminished. There are four such formulae, as follows: 



INTEGRATION BY PARTS. REDUCTION FORMULAE 285 

(i\p + m + l)b (np + m + l)bJ K } ' K J 

Cx m (a -f bx n ) p dx 

= ar»(a + bary *pa Tarfr + toy^ifr . . . . (B) 

np + m-\-l np + m + U 

Cx"\a + bx n ) p dx 

_ x^ja + bx")^ 1 _ (»j) + « + m + l)5 C m+n , b yd (c 

~ ( m + l )a (ro + l)a J * & + *")«*' ( C ) 

r.r m (a + ^) p ^ 

= _ ^^^ ¥ + M+w + l ^ ( ^ lfe ( 

n(p + l)a n(jp + l)a J K J v J 

Formulae (A) and (5) are used when the exponent to be reduced, 
m or p, is positive, (^L) changing m into ?ti — w, aud (B) changing p 
intop — 1. 

Formulae (C) and (D) are used when the exponent to be reduced, 
m or p. is negative, ((7) changing m into ?/i + ?i, and (D) changing p 
into p + 1. 

If, in the application of one of these formulae to a particular case, 
any denominator becomes zero, the formula is then inapplicable. 
For this reason, 

Formulae (A) and (B) fail, when np + m + 1 = 0. 
Formula (C) fails, when m + 1 = 0. 

Formula (D) fails, when p -f- 1 = 0. 

In these exceptionable case's the required integral can be obtained 
without the use of reduction formulae. 



286 INTEGRAL CALCULUS 

219. Derivation of Formula (A). Let us put for brevity 
X = a + bx n , dX = nbx n ~ l dx. 



Then Cx m X p dx = Cx n 



-n+l 



X p dX 
nb 



Integrating by parts with u = se m-n+1 , we have 

Cx m X p dx = xm ~ n+lXP+1 _ m 7 yi + 1 Cx m ~ n X>+ 1 dx. . . (1) 
J 7ib(p + l) nb(p + l)J K J 

Comparing the integrals in (1), we see that not only is m diminished 
by n, but p is increased by 1. 

In order that p may remain unchanged, further transformation is 
necessary. 

By substituting X p ^ = (af bx n ) X p , 
the last integral may be separated into two. 

Cx m ~ n X p+1 dx = a Cx m - n X p dx-\-b Cx m X p dx. 

Substituting this in (1) and freeing from fractions, 
nb(p + 1) Cx m X p dx = x m ~ n+1 X p+1 

— (m — n + l)fa Cx m ~ n X p dx + bfx X p dx\ 

Transposing the last integral to the first number, 

(iip + m + 1) b Cx m X p dx = x m ~ n+1 X p+1 — (m — n + l)a Cx m ~ n X p dx, (2) 

which immediately gives (A). 



INTEGRATION BY PARTS. REDUCTION FORMULA 287 

220. Derivation of Formula (B). Integrating by parts with 
u = X p , we have 

( %-X* dx = X*^—-- I — pX^bnx"- 1 dx 

J m + 1 J m + 1 

= ^^_m>br xm+nXP - ldx ^ . . . (1) 

m + 1 m + lJ J 

Comparing the integrals, we see that not only is p decreased by 1, 
but that m is increased by n. 

To avoid the change in m, substitute in the last integral of (1) 

bx n =X-a. 

Also freeing from fractions, 

(m + 1) Cx m X p dx = x m+l X p - npf Cx m X p dx-a Cx m X p ~ l dx). 

Transposing to the first member the last integral but one, 

(np + m + 1) Cx m X p dx = x m+l X p + npa Cx m X p ~ 1 dx, ... (2) 
which immediately gives (B). 

221. Derivation of Formula ((7). This may be obtained from (2), 
Art. 219, by transposing the two integrals, and replacing throughout, 
m — n by m. This gives 

(m + 1) a Cx m X p dx = af l+1 X^ p+1 — (np + m + n + 1) Cx m+n X p dx, 

from which we obtain (C). 

222. Derivation of Formula (D). This may be obtained from (2), 
Art. 220, by transposing the two integrals, and replacing p— 1 by p. 
This gives 

n(p + l)a f afX* cte = - x m+1 X p+1 + (np + n + m + 1) Cx m X p+1 dx, 
from which we obtain (Z>). 



288 INTEGRAL CALCULUS 



EXAMPLES 

Here f ^ da; = Cx\a 2 -x 2 )~^dx. 
J Va 2 — sc 2 ^ 

Apply (A), making 

m = 2, n = 2, p = — -, a = a 2 , &= — 1. 



fx\a 2 - x 2 y* dx = < a *-f)* _ J^ f( a 2 _ x yi dx 
«/ — ^ — zu 



= (a 2 — x 2 )* 4- — sin -1 -. 

2 V ' 2 a 



2. fV<* 2 + a 2 cte = | Vtf + a^ + ^logfc + V« 2 + sc 2 ). 
»/ Z Z 

Apply (jB), making 

m = 0, w = 2, p = -, a = a 2 , 6 = 1. 

z 



JV + a0**« = f (*'+af)* + f /■ 



da? 



(a 2 + x 2 y 



*( a 2 + a?)i + 1 log(z + V^+^). 



/die V#— a- . 1 ,0? 

ar V# 2 — a 2 ^ ax " ^ a a 



INTEGRATION BY PARTS. REDUCTION FORMULAE 289 

Apply (C), making 

m = — 3, n = 2, p = — -, a= — a 2 , b = 1. 



JV^ - tff**- ^~ rf >* + ^/^(^ - *>-»* 



2 aW 2o 8 a* 



,. r dz 1 1 _,» 

*j l — - . -S6C 



Apply (D), making 

3 
m = — 1, ?i = 2, p = — -, a = — a 2 , 6 = 1. 

«/ CI QT*J 



1 _iiC 

.see 1 - 



a 2(^_ a 2)i a 3 a 



5. C^Ja 2 -x 2 dx = -^a 2 -x 2 + - sin" 1 1' . 
J 2 2 a 

J VflS? — a 2 ^ ^ 

7. f(a 2 - a 2 )* <fo = -(5 a 2 - 2 x 2 ) V^^ 2 + ^ sin_1 -• 
»/ 8 8 a 

8. (V-a?)$ <fe = *(2 x 2 - 5 a 2 )V?=7+^ 4 log (a? + VaF+d). 
*J 8 8 



290 INTEGRAL CALCULUS 

9. fx 2 ^/a 2 -x 2 dx = ^(2 x 2 - a 2 ) Va 2 - x 2 + - sin" 1 -• 

10. fxWx^+ti 2 dx = ^(2x 2 + a 2 ) ^/x T ~+a~ 2 - ^ log (a + V^T^" 2 ). 
»/ o 8 



C?iC 



(a 2 -a 2 )t a 2 Va 2 -aJ* 

12. Derive the formula of reduction used in Case IV of Rational 
Fractions. 

/ OjX J. X . /(~\ Q\ I U'37 

J (x 2 + a 2 ) n = 2(n-l) a 2 [_(x 2 + a 2 )*- 1 l "~ V (x 3 +a 2 ) n ~ 1 _ 
1Q f da; 3 x 2 — a 2 



sec 



is. r_ ^jx_ = _ ^ v __ 5 + 3^ ^^ 

^ V 2 aa? — or * 2 a 

Write P ^^ |" s'cfa ? and apply (^) twice 

•^ x V2 asc — x 2 J V2 a — x 



*g /* dx V 2 a# — x 2 . 

*^ x\/2ax—a?dx aiC 



17. f V2 az - x 2 dx = ^* V2 ax - x 3 + f sin" 1 5= 
^ 2 2 a 



or = — - — V2 a# — # 2 + — vers -1 
2 2 a 



INTEGRATION BY PARTS. REDUCTION FORMULA 291 

Write J V2 ax — a* 2 d.r = I Va 2 — (x — a) 2 rife, and substitute 
n Ex. o. 



18. fx V2^T? fe = _ o «- + ax - 1 *■ V o^r^ 2 + g 
«y o ... 

1ft f* V2 ooj — x 2 dx /« \ , -i#. 

19. I = V2 ax — x- + a vers * - ■ 

»/ a; a 



3 ,3! 

- vers -1 -. 

2 a 



20 f ' 1 '"' r? ' 1 ' — - a- 7 "- 1 V2 aa — it- 2 (2 m — l)a /» a;™- 1 eta 
J a 2 a.i- - .r ~ »» w» ^ V2 aa; - or 5 * 

21. f ** 

*^ x m V2 ax — x 2 



V 2 ax— x 2 , m — 1 /• cfa 



+ 



- l)a J i 



(2 m - l)ax m (2 m - l)a J x m ~^2 ax - ar 2 
22 . | x m V2 aa; - x 2 dx 

m + 2 m + 2 •/ 

23 t f i'- aa: — x 2 dx 



(2 aa; - x 2 )* m-3 r ^2ax-x 2 dx . 

(2 m - 3)aa;" (2 ra - 3)aJ a;" 1 " 1 

223 Trigonometric Reduction Formulae. — The methods explained 
q Arts. 211, 214 are applicable only in certain cases. 
By means of the following formulae, 

I sin m x cos n x dx, J tan" 1 x sec n x dx, and J cot m x cosec n x dx, 

lay be obtained for all integral values of m and n> by successive 
eduction. 



sin TO x cos n x dx 



292 INTEGRAL CALCULUS 

f sin" x cos" x dx = - slnW-1 ^ CQS " +1 ^ + *2?^l1 fsin— 2 aj cos w x dx. (1) 
J m + n ra + nJ 

Psin" a; cos" x dx = smm ^ x cosW " 1 x + -^— ^- f sin™ a cos w - 2 a da;. . (2) 
*/ m + n m + n*/ 

/ 

JS»J«3» + Si!±!f l ii-., M .,fc . . . (3) 
m + 1 m-\-l J 

I sin" 1 a? cos w x dx 

sin m+1 x cos w+1 x . m-\-n-\-2 C - m n+2 ^ sa\ 

= ' — I sm m x cos n+J x dx. . . (4) 

71 + 1 71+1 J 

J. _ , sin m_1 x cos a? . m — 1 C ■ m _ 2 -. /KN 

sm m xdx = 1 I sin™ 2 xdx (5) 
m m J 

/_ , sina;cos M_1 a? , n — 1 T n _ 2 , /a , 

cos w x dx = h — I cos w 2 x dx (o) 
7i n J 

f tan™ a? sec' 1 x dx = tanW ~ 1 ,T sec " x m ~ 1 ("tan" 1 " 2 x see" a? da;. (7) 

J m + n — l m + n — U 

I cot m a; 

_ £2 

m + 7i — 1 m + n 

/-, sec n_2 x tan a; . n — 2 C „_ 2 , / > 

sec n a; da; = 1 I sec w2 a; da; (9 
n — 1 7i — 1*/ 



cosec w a? da? 



cot m_1 a; cosec" x m — lC ±m-2 „,>„«„* „ ,7™ /q> 

- I cot m z x cosec n x dx. . \6 J 



INTEGRATION BY PARTS. REDUCTION FORMULAE 293 

cosec 1 x dx = 1 I cosec n ~- xdx. . . . (10) 

n — 1 n — U y 

Ctmi"xdx = tan "~ lx - Cteti*-*xdx (11) 

Ccot n xdx = - Qotn ,v - Ceot n - 2 xdx (12) 

224. Derivation of the Preceding Formulae. — To derive (1), we 
integrate by parts with u = sin m_1 x. 

f sin- x cos" x dx = - sm "" 1 x cos " +1 x + 7 Azil f sin— 2 x cos n+2 x dx. 
J n + 1 n-\-lJ 

I sin- -2 x cos"^ 2 as dx = J sin" 1-2 a; cos" x dx — J sin w x cos n x cfa. 

Substituting this in the preceding equation, and freeing from 
fractions, we have 

(m + ») I sin m a; cos' 1 a; cto 

L m_1 » cos" +1 # + (m — 1) J sin TO_2 a? cos" a? cfa, 



= — sin" 



which gives (1). 

To derive (2), integrate by parts with u = cos 71 " 1 x, and proceed as 
in the derivation of (1). 

Formula (3) may be derived from (1) by transposing the integrals, 
and replacing m — 2 by m. 

Formula (4) may be derived from (2) by transposing the integrals, 
and replacing n — 2 by n. 

To derive (5), make ra = in (1); and to derive (6), make m = 

The derivation of (7), (8), (9), and (10) is left to the student. We 
have already derived (11) and (12) in Art. 212. 



294 INTEGRAL CALCULUS 

EXAMPLES 

n C • 6 7 cos x /sin 5 x . 5 . » . 5 . \ , 5x 

n C 5 7 cos xf 1 , 3 \ . 3i , X 

2. I cosec 5 xdx = -— - — f- . „ + t: log tan-- 

J 4 \sm*x 2sin 2 xJ 8 B 2 

o /• 7 , sin x f 1 , 5 . 5 N 

3. I sec 7 x dx = — — | — + - 

J 2cos 2 oA3cos 4 # 12 cos- a? 8, 



5 

+ — log (sec x + tan x). 



A C * a sina/ - ,7 5 ,35 3 ,35 \ , 35a; 

4. I cos 8 #cta = — — cos 7 a + -cos 5 # + — cos 3 # +— cos#+tt— • 
J 8 \ 6 24 16 y 128 

K /* • 4 o ^ cos # /sin 5 x sin 3 ic sinaA . a; 

5. I sm 4 x cos 2 x dx = ^ f — - ] + - — 

J 2 ^ 3 12 8 J 16 

c f % cos*x 1 3cosa? — 4 cos 3 a; 3coscc , 3, , x 

6 -r-y-*8= -p- — A o • 2 + g log tan-- 

J snr x 4 sm 4 x 8 sir a; 8 2 

7. J-^^-^^+^-lsin; 



sin 4 a; cos 3 x cos 2 # V 3 sin 3 x 3 sin a; 2 



5 
+ - log (sec x + tan #). 



A 4 o , /tan 3 a; tanaA 3 , sec a? tan # 

8. I tan 4 sc sec 3 a cte = -— sec 3 aH • 

J \ 6 8 / 16 

+ — log (sec x + tan x). 

9. rcot^cosec 5 ^^^ 60 ^ 0086 ^^- 60566 ^^ 008602 ^^-- 
J 2 V 3 12 8 

-ilogtan| 



CHAPTER XXVI 
INTEGRATION BY SUBSTITUTION 

225. The substitution of a new variable has been used in Chapter 
XXIII, for the rationalization of certain irrational integrals. We 
shall consider in this chapter some other cases where, by a change 
of variable, a given integral may be made to depend upon a new 
variable of simpler form. 

We shall first consider some substitutions applicable to integrals 
of algebraic functions, and afterward those applicable to integrals 
of trigonometric functions. 

226. Integrals of form I f(a?) xdx, containing (a + bzr)q. One of 
the most obvious substitutions, when applicable, is x 2 = z. 

By this, any integral of the form ifix^xdx 

is changed into - \f{z)dz. 

p 
Integrals containing (a + 6.x* 2 )? are often of this form. 

x 3 dx 



Take for example - J - 



VI -x 2 
By the substitution x 2 = z, 

C x'dx _ 1 r zdz 

This is of the form of Art. 203, and is rationalized by putting 
l-z = v:\ 

295 



296 INTEGRAL CALCULUS 

The two substitutions in succession are equivalent to the single 
substitution 1 — x 2 = id 2 . 

Applying this to the given integral, 

x 2 = 1 — w 2 , xdx = — w dw. 

f ^ dx =- r a-"')wdw = _ r (X _ w r )dw 
J vi - « 2 J w J 



= -^_| 3 ) = -|(3-^) = -^^(^ + 2) 



EXAMPLES 



J V2s*+1 30 

2. far* (a 2 - x 2 )^dx = -£-(6 a 4 - aV - 5 a 4 )(a 2 - x 2 )*. 



3. C dx = 1 log ^' + ^-^ ^-Llog «* 

JaVa 2 + a 2 2a Va 2 + a 2 + a 2a (yV + a 2 + a) 2 

1 t aj 

a Var 2 + a 2 + a 

4 . r °g? = |r^±i)! + (^ + ,i)i + iog(^+i-i)' 

•^ v x 2 + 1 — l ^ L ^ 

5. f xdx * log ( V3~=^ + 1) + j log ( V3^¥ 2 -3). 

^ a^ + 2V3 - OJ 2 4 4 



227. Integration of Expressions containing Va 2 — ar 2 or V&* 2 ± a 2 , 
by a Trigonometric Substitution. Frequently the shortest method of 
treating such integrals is to change the variable as follows : 



INTEGRATION BY SUBSTITUTION 297 



For a a* — .r , let as = a sin or x = a cos 0. 



For vr + o-, let x = a tail or a; = a cot 0. 
For v .r — a 2 , let # = a sec or a* = a cosec 0. 

Jdv 
. 
(a 2 - a,- 2 )* 

Let x = a sin 0, cfcc = a cos dO, 

cr — or = a- — a 2 sm- = <r cos- 0. 

cos Ode 1 C cJO tan (9 



/ c7x _ a, cos e do __ i r 
(a* — a?)% ^ a 3 cos 3 a 2 J cos 2 6 a 2 a 2 Vd 2 — x 2 

nple j — 

J x^Jx 2 + 



Take for another exai~~ 



Let x = a tan 0. 



/ dx _ r a sec 2 dO __ 1 /" sec d n_l C dO 
x^/x 2 + a 2 ^ a tan • a sec a J tan a J sin 



- log (cosec - cot 0) = 1 log Vx 2 +a 2 -a 
a a x 



Again, find I ^ ' 

Let x = a sec 0. 



^-« \te 



J a; J a sec J 

= a f (sec 2 - l)d0 = a (tan 0-0) 



Vx 2 — a 2 — a sec -1 - • 
a 



298 INTEGRAL CALCULUS 



EXAMPLES 

,2 



1. I Va 2 — x 2 dx = - Va 2 — x 2 -j — sin -1 - • 
J 2 2 a 

n C dx _ _ Va^ + a 2 
J a; 2 Va?Ta 2 ^ 

3. f dx = \og(x4-Vx T ^~a 1 \ 

J ^Jx 2 — a 2 



4 C dx = (2a; 2 + a 2 )Va 2 -^ 2 . 

5. I — — ^— dx — — — — -L 1_ log (x + Var -h a 2 )- 

J ctr rr. 



J x 4 3a 2 x s 



7 C_dx__ =z (2x 2 -l)Vx 2 + l 
J x'Vx^+l ~ 3x* 

8. fv/^±i da? = f * + 1 da; = V^I + log (x + V^l). 



9. r — g* — =\/^- 

^ (a3 + l)V^ 2 -l ^x + 1 

io. f *? =J5±i. 

11. f *<* g == _V 8 + 2cc-a; 2 + sm-^ 

^V3 + 2x-aj 2 2 



INTEGRATION BY SUBSTITUTION 299 



For V3 4- 2 x - x* = V4— (x— l) 2 , let x — 1 = 2 sin 0. 

r <f.r 

J (j* + 2x + 3)* 



12. " + 1 



2 Va 2 + 2 a; + 3 



For Var>+2 x + 3 = V(<e+ 1) 2 + % let a; + 1 = V2 tan 0. 

13 C r *' v — x — a 

J (2ax-x*)% aW2ax-x* 

UC ja — x _ C a — x , / s , « • _i2.T-ffl 
. I \i da; = I — ^=r cfo = wax—x- + - sin l 

J V * ^ Vaaj-ai 2 2 a 

15 (* a ' 2 ^ x (3a-\-x) V2 aa; — x 2 3a 2 . _! «; — a 

' J V2^"T~'~ 2 ~ 2 Sm a " 

228. Substitutions for the Integration of Trigonometric Functions. 

A trigonometric function can often be integrated by transforming it, 
by a change of variable, into an algebraic function. For this purpose 
two methods of substitution may be used, as shown in the two follow- 
ing articles. 

229. Substitution, sin x=z, cos x = z, or tan x = z. 

Consider, for example, C sin x cos x dx 

J 1 — sin x + cos 2 x ' 

dz 

Let sin x = z, then x = sin -1 z, dx = 



VI -z 



J sin x cos x dx _ C z Vl — z 2 dz _ C zdz 

1 — sin x + cos 2 x J 1 — z + l—z 2 ^/\ _ 2 2 J 2 — z — z 2 

C zdz = 2 r dz 1 r_dz_ 

J (2 + z)(l-z) 3 J 2 + 2 3 J 1-2 

= - ? log (2 + 2) - I log (1 - 2) = - | log [(2 + sin a;) 2 (l - sin «)]. 
o o o 



300 INTEGRAL CALCULUS 

EXAMPLES 

1. C f , =-1— r^-^taii-^tan^l 

Ja 2 +b 2 t3in 2 x a 2 -b 2 [_ a \a )] 

2 r — dx — = x i j , in cos v 

J 1 + tanx 2 2 6 v ; 

3. f^^^ = lio g l-sm^ + J_ log l + V2sinx > 
J sin 4 a? 8 1 + sinsc 4V2 1 — V2sina; 

Let sin a; = 2. 



dx 1 n sin a; (1 + cos x) T , 

= - log i — ! '-. Let cos a; = z. 

sin2x 3 5 (l+2cosa;) 2 



4. f ° 

»/ sinaj + 

K /• sin x + cos x 7 3 oj 1 1 • ■ l0 \ 

5. I — dx = log (sin x + 2 cos x) 

J sin a? + 2 cos a; 5 5 



Let tan x = z. 

/'tan 3 x , , 1 i 
6. I dx = x -\ = loj 



'tan 3 a;-, . 1 , tan x — V 3 

dx = x-\ log -. 

tan x ^3 tan x + ^J§ 



7. Show, by transforming into algebraic functions, that only one of 
the following integrals can be expressed in terms of the elementary 
functions. (See Art. 208.) 



/Vtana?d!a;= Cxi!*?, where z = tana;. 

J 1 + z 2 

I Vsina?daj= I — = | — , where z — sin x. 

J J Vl-z 2 J Vz-z 3 



INTEGRATION BY SUBSTITUTION 301 



230. The Rational Substitution, tan - = z. By this substitution, 
sin x, cos x, tan x, and dx are expressed rationally in terms of z. For 



siu x - 


2 tan - 
2 


22 


COS X- 


1 + tan-- 

1-tan 2 * 

2 


1-Z 2 




1+tan 2 - 
2 


~1+Z 2 



2 tan - 

9 



tana; 



1-tan 2 * i-* 2 

2 



From - = tan- 1 2, dx = ^- c 



9. 



It follows that the integral of any trigonometric function of x, 
not containing radicals, may be made to depend upon the integral 
of a rational function of z, and can therefore be expressed in terms 
of elementary functions of x. 

C dx 

231. To find | Applying the substitution of the 

J a + bsmx ™ J s 

preceding article, tan - = z, 

2dz 

2dz 



/i z riz 
i + * 2 r 
, 2bz J a 
& ~r z « 
1 + 2 2 



(l + *) + 2ki 

+ ' 



/ 2fttfe _ /" 2 r/rfc 

a 2 z 2 + 2abz + a* J (az + b) 2 + a 2 - 



b 2 



302 INTEGRAL CALCULUS 

If a > by numerically, 



x 

a tan - + b 



r_dx__ = 2 t ^_ Y az + b = 2 ^^ 2 t 

J a + 6 sin a; Va 2 -6 2 Va 2 -6 2 Va 2 -6 2 Va 2 -6 2 



If a < 6, numerically, 



/ da? _ r 
a + 6 sin cc J (i 



2a& 



log a^ -6-V6 ; 



a Z + &)2_(52_ a 2) y 6 2_ a 2 a2 + &+ V & 2_ a; 



a tan - + 6 — V& 2 — a 2 

log 

atan| + & + V& 2 -a 2 



V& 2 - a : 



232. To find 



f- 

J a 



dx 



-j- b COS £C 



/ 2 dz 
6(1— z 2 ) ~~J ( 



2^2 



a — b) z 2 + a -f & 



2 /" dz_ 



+ &'' 



a — b 



f- 

J a 



If a > b, numerically, 

dx 
+ b COS £C 



a-6^a+6 Va+6 



_2 tan-'f J«L^ tan 



INTEGRATION BY SUBSTITUTION 303 

If a < b, numerically, 

dx 2 r dz 



r dx = 2_ r az 

J a + b cos x b — aJ «a _ b + a 



b — a 



1 1 gv6-g-V& + a 
V& 2 - a 2 sV&-a+V& + a 



.r 



Vfr — a tan h + V b + a 
log 



V& 2 — a 2 / # / 

V 6 — a tan « — V & + a 



EXAMPLES 

Integrate the following functions by means of the rational sub- 
stitution. 

1. J,-^- = W^2tan* 



3cosa? 2 V 2 , 

2 f (? * = 1 j tanaj + 2-V3 

" J l + 2sin2a 2V3 & tana; + 2 + V3 

3tan^ 

3. f * = i log I 

J 5 sin x + 12 cos a 13 & , x 



in^ + 2 



2 tan ^-3 



&tan£-a + Va 2 + & 2 



J a sin* + 6 cob* V^+6 5 ° g 6 taD * _ a _ V^+6 5 

5. f * = log- 

J sin ./,* -f- vers as ^ 



tan 2 



x 



H- tan - 

2 



304 INTEGRAL CALCULUS 

tan--l 

6. f *» = tan-_J? 

J 3 — sm x + 2 cos a? 2 

'• I 1= — ^ = ^log 

Jo + T sm a; — cos x 5 



3tan| + l 



tan^ + 2 



/dx 1.x 1 i A 

— . . = - tan — log f 1 + 
(1 + sina + cosaj) 2 2 2 1+tan » V 



tan 



233. Miscellaneous Substitutions. Various substitutions applicable 
to certain cases will be suggested by experience. 

The reciprocal substitution, x = ~, may be mentioned as simplify- 

z 

ing many integrals. 

EXAMPLES 
Apply trie reciprocal substitution x = - to Exs. 1-6. 



L J x 4 3aV 



2 r___dx__ _ _ Va?-|-a 2 
' J fvV + a 2 " a^» 

3 r V2 ax - a; 2 ^ = (2 ax - x 2 )? ^ 
J x s Sax 3 

4. r ** =ii og g 

J j;VV ± x 2 a « + Va 2 ± x 1 - 



5 . r ^-f)* ^ 

J x 4 



3JX-X 3 ) 
8tf 



4 

ill, 

,4 



INTEGRATION BY SUBSTITUTION 



305 



r *b = 

J ±\ 8 x 2 + 2 x - 1 



snr 



x-1 
3x 



7. f "*** = 5 §_ + _6_ + i g( a . + 2). 

Let a; + 2= 2. 



8 . r ix+a)(x+brdx= ^f +(a _ b) (^i 

J n -\- Z 91 + 1 



Let x-\-b — z. 



J x (a + &a?) 3 a 3 _2(a + &a>) 2 a + bx x 



Let a + 6a; = a;z. 



10. - dx = sin a; — sin a log tan * "T • Let a; + a = 2. 



tan (a; + a) 



1 1 . f - — ^ = - a? - «- + log Ve & + e* + 1 — tan 



1 ^.,2^ + 1 



V3 V3 

Let e x = z. 



12. f\- — -dx = V(a - a)(& + a) + (a + 6) sin" 1 J- 
J ^ b 4-x * a 



+ b 



+ 

Substitute b + x = z 2 , and the integral takes the form of Ex. 5, 
Art. 222. 



13. 






= V(a; + a)(x + b) + (a- 6)log(Va; + a + Va; + 6). 



Substitute x + 6 = z 2 , and the integral takes the form of Ex. 2, 
Art. 222. 



CHAPTER XXVII 



INTEGRATION AS A SUMMATION. DEFINITE INTEGRAL 



234. Integral the Limit of a Sum. An integral may be regarded 
and defined as the limit of a sum of a series of terms, and it is in 
this form that integration is most readily applied to practical 
problems. 

235. Area of curve the limit of a sum of rectangles. Let it be re- 
quired to find the area PABQ included between the given curve OS, 
the axis of X, and the ordi- 
nate s AP and BQ. 

Let y=x? be the equa- 
tion of the given curve. 

Let OA = a, and OB=b. 

Suppose AB divided into 
n equal parts (in the figure, 
n = 6), and let Ax denote 
one of the equal parts, AA 1} 
A\A 2 , ••• 

Then AB = b — a = n Ax. 

At A lf Ac,, •••, draw the 
ordinates A X P X , A 2 P 2 , •■•, and complete the rectangles PA X , P Y A 2 , 

From the equation of the curve y = x?, 

PA = a\ P X A X =(a + Ax% P 2 A 2 = (a + 2 Ax)?, •••, QB= b?. 

Area of rectangle PA 1 — PA x AA X = a? Ax. 

Area of rectangle P±A 2 = P^A X x A X A 2 = (a + Ax)? Ax. 

Area of rectangle P 2 A 3 = P 2 A 2 x A 2 A 3 = (a + 2 Ax)? Ax. 

Area of rectangle P 5 B = P 5 A 5 x A B = (b — Ax) ? Ax. 

306 




INTEGRATION AS A SUMMATION. 307 

The sum of all the n rectangles is 

a 1 - A.r -f (a + Aac)* A.r -f (a + 2 A.r)*- As H h (& — As)* Aa?, 

^ 5 i 
which may be represented by > .r-' A.r, 

l a 

where x- A.r represents each term of the series, x taking in succes- 
sion the values a, a + A.r, a + 2 A.r, •••, b — A.r. 

It is evident that the area PABQ is the limit of the sum of the 
rectangles, as n increases, and A.r decreases. 

That is, Area PABQ = Lim Ax=0 > x°- Ax. 

' a 

236. Definite Integral. From the preceding article 

V x* A.r = a* A.r + (a 4- Aa)* Ax'+ (a -f 2 Ax) * Ax H h (6 — Ax) i A.r. 

The limits of this sum, as A.r approaches zero, is denoted by 
J .r- dx. That is, by definition, 



| «2 dx = Lim A:e=0 > & Ax. 

a ^ a 



X h 1 1 

x- f?.r is calleji the definite integral, from a to 6, of a^da?. 

It is to be noticed that a new definition is thus given to the sym- 
bol I . which has been previously defined as an anti-differential. 

The relation between these two definitions will be shown in the fol- 
lowing article. 

237. Evaluation of the Definite Integral I x% dx. This is effected 
by finding a function whose derivative is a?* 

dx\ 3 J 
By the definition of derivative, Art. 15, 

, - (a -f Ax) 2 

— f ] = Lim . = s*. 

da\ 3 J ^ A,; 



308 INTEGRAL CALCULUS 

T = X* + C, 

Ax 

where e is a quantity that vanishes with Ax. 

2 3 2x* i 
Hence -(x4-Ax) 2 ^— =x 2 Ax + eAx. 

o o 

Substituting in this equation successively for x, 
a, a-\- Ax, a + 2 Ax, •••,& — Ax, 

| (a + Ax)* - ^ = a* Ace + <a Ax, 
o o 

^ (a + 2 A a)* — - (a + Ax)^ = (a + Ax)? Ax + e 2 Ax, 
o o 

| (a + 3 Ax) * - ? (a 4 2 Ax)* = (a + 2 Ax)* Ax + € 3 Ax, 
o o 



. 26f _ | (ft _ Ax) 3 = (6 - Ax)?Ax + e n Ax. 
o o 

Adding and cancelling terms in the first members, 

?A 2 _ 1^1 = a i Ax + (a + Ax)*Ax 4- (a 4- 2 Ax)? Ax H 1- (6 - Ax)?Ax 

+ e 1 Ax + e 2 Ax 4 e 3 Ax 4 ••• -f e M Ax 

= V^Ax + Y^Ax (1) 

^^ a a 

Comparing with the figure, Art. 235, X x 2 Ax, as we have before 
seen, represents the sum of the rectangles, and 2j e Ax represents 

a 

the sum of the triangular-shaped areas between these rectangles and 
the curve. 

The latter sum approaches the limit zero, as Ax approaches zero. 



INTEGRATION AS A SUMMATION 809 

For if e A . is the greatest of the quantities c 1} u, •••£„, it follows that 
V c Xv < e,T Ax, 

1 a ^^ a 

that is, >, e & x < e /t(^ — a )- 

As e t vanishes with Ax, 

Lim ^^ eAaj = 0. 



Taking the limit of (1), 

a?s da; = ^iL-— = Area PABQ. 

Thus the value of I x* dac is found from the integral 

S xldx = 2 4> 

by substituting for a;, b and a in succession, thus giving 

3 3 

The process may be expressed 

I 6 2b% 2 J 



(Mo.-** 

Ja 3 



3 



This is called integrating between limits, the initial value a of the 
variable being the lower limit, and the final value b the upper limit. 
In contradistinction 

f x l d x = 2 -^+C, 
is called the indefinite integral of xMx. 



310 INTEGRAL CALCULUS 

238. General Definition of Definite Integral. In general if f(x) is a 
given function of x which is continuous from a to b, inclusive, the 
definite integral 

J / (x)dx = Lim Aa . = o /(a)Aa> + f(a + Aa) Ax + /(a + 2 Ax)Ax 

+ - +/(&-Ax)Ax . 

If j f(x) dx = F(x), the indefinite integral, 

l f(x)dx = F(b)-F(a) --.(I) 



This may be illustrated by the area bounded by a curve as in Art. 
235, by supposing y—f(x) to be the equation of the curve OS. 

The proof of Art. 237 may be similarly generalized by substitut- 
ing /(» for a?*, and F(x) for ~- 

Geometrically the definite integral I f(x)dx denotes the area 

swept over by the ordinate of a point of the curve y =f(x), as x 
varies from b to a. 

It is to be noticed that in Art. 192, by a somewhat different course 
of reasoning, we have arrived at the same result, 

Area PABQ = F(b) - F(a). 

239. Constant of Integration. It is to be noticed that the arbitrary 
constant G in the indefinite integral disappears in the definite 
integral. 

Thus, if in evaluating J x* dx, we take for the indefinite integral 



/ 



3 ' 



2 6* 2 J 



wefind jyUx = *f + C-(?f+c)== 3 g 

Or if Cf (x) dx = F (x) + C, 

Cf(x)dx = F(b) + G- IF (a) + C]=F.(b)-F(a). 



INTEGRATION AS A SUMMATION 



311 



EXAMPLES 

.0 

1. Compute 2 -^Ax for different values of Ax. 
When Ax = .2, 

X/ Ax = (l 2 + L2 2 + 1A 2 + D? 

+ r8 2 )(.2) = 2.01. 
When Ax = .1, 
2^x 2 Ax = (I s + 13* + V? + • • - 

+ L9 2 )(.1) = 2.18. 

When A.? = .05, Jj'asPAa = 2 - 26 - 
Lirn Ax=0 5j a~ Ax = ( x 2 c?x = — 



2.33. 



Curve OS, y=x*. 0A=1, 0B=2. o 
Area PABQ = 2.33 square units. 




2. Compute V — for different values of Ax. 

*"*1 x 



'Ax 
x 

When Ax = 1, 
4 Ax (\ . 1 , 



Y |R 



s.T-e+i+i^- 1 "- 



When Ax = .5, 



Ax 



^A ■ 1XX _ 
AT" 



1.593. 



When Ax 



Ax 



^1 X 



1.426. 




312 INTEGRAL CALCULUS 



Lim Aa;=0 > — = — = log x 

**l X Jl X 



= log 4 - log 1 = log 4 = 1.386. 



i 



Curve RQ, y = -> 0A = 1, 0B = ±. Area PABQ = 1.386 square 
units. 



3. Compute Y x Ax, when Ax = 1 ; when Ax = .5 ; when Ax = .2. 
3 ^Ins. 18; 19; 19.6. 

Find Lim A3;= o 2y ^ Acc - ^ ns - 20 - 



4. Computed -, when Ax =.2; when Ax = .l; when 

Ax =.05. ol + x Am .833; .810; .798. 

Find Lim Ax=0 Y* -**L. Ans. £ = .785. 

^o 1 + x 2 4 



5. Compute V log 10 xAx, when Ax = l; when Ax = .5; when 
Ax =.3. 10 ' Ans. 3.121; 3.150; 3.161. 

Find Lim Aa;=0 > log 10 xAx. Ans. 13 log 10 13 — 3 log 10 e — 10=3.177. 



6. Compute V tan<£ A<£, when A<£ =3°=^; whenA<£ = -^-; when 

^^ IT ou oU 



A<£ = -^-. 
V 180 



125 
6 ' 



Find Lim AcE=0 V tan <f> A<£ 

4 

7. r 3 (ic 2 -4) 2 x<!x = 

- ; « Va 2 -2/ 2 3 

Jo e 2 * + l 4 



.4ns. .316; .328; 


.340. 


Ans. log e V2 = 


.346. 


/* 9 x dx _ 9 




J 5 Vx 2 + 144 




r 2 dx * 

Ji x 2 -x + l 3V3 









INTEGRATION AS A SUMMATION 313 

13. C flv - = ff . r f 8 

15. jTsin^ < M = ^_|. 16 _ r vers22e ^ = ^_7V3. 

is «/o 4 16 



"f. 



,; cos 20 — cos 2 a 1A -, , tt 



cos 6 1 — cos a 



dO = 1 + - cos a. 



18. ■= 6 log 2. 

Jo (aj + 2) 3 4 & 

19. f 1 — = -log (2 e). 

Jo l + 2x + 2x 2 + 2x s + x* 4 & ^ ; 

7T 

20. .i*- sin .r dx = 7r — 2. 

21. f 2a a log (x + a) cfc = ^log (3 a) - 1. 

%J a A 4 

22. f'tan- 1 -dx = tt - log 4. 

•/o 4 

By (5) and (6), Art. 223, we find 

J sin" ;/: dx = ?/ ~ ( sin n ~ 2 a; cfo, 
o n Jo 

7T 7T 

I cos n x ax = I cos n-_ x ax ; 

Jo n Jo 

from which derive the following results : 



814 INTEGRAL CALCULUS 

24. If n is even, 

E E 

J"\in n xdx = C\os> n xdx = 1 ' S ' b '"( n ~ V ) 
o Jo 2-4-6.--H 



25. If n is odd, 

E E 

J" 2 C 2 2 »4. 6 •••(71 — D 
sin" x dx = I cos M # d# = ^ £• 
o c/o 3 • 5 • 7 • • • n 

240. Sign of Definite Integral. In considering the definite inte- 
gral I / (x) dx, we have supposed a < 6, and / (x) to be positive be- 
tween the limits a and b. 

Iif(x) is negative from x— a to x = b, V /(a?) A#, being the sum 

of a series of negative terms, is negative, and consequently 

Jf (x) dx is negative. 

If /(a;) changes sign between x = a and x=b, I /(#) cto is the 
algebraic sum of a positive and a negative quantity. 
Y 




For exa 



mple, £ 



I 



cos x dx — 1 = area ^4 0J5. 



cos x dx == — 2 = area JBCZ). 



INTEGRATION AS A SUMMATION 315 



i 



3ir 

2 

cos x dx = —1 = 1 — 2. 





f """ cos x dx=0 = l — 2 + 1. 
The change of sign resulting from a < b is considered in Art. 243. 

241. Infinite Limits. In the definition of a definite integral the 
limits have been, assumed to be finite. When one of the limits is 
infinite, the expression may be thus defined : 

J fix) dx = Lim 6=w \ fix) dx. 

For example, consider Ex. 12, following Art. 239. 

r~8a s dx T . r b Sa s dx T . A ,, _ x b\ 2 

o , , o = Li m 5=» I o , , o = Lnn 6=ao 4 ^ tan x — = 2tt a 2 . 
Jo ar + 4cr Jo or + 4 a 2 \ 2a J 

Keferring to Art. 126, we find the geometrical interpretation of 
this result. 

The area included between the curve, the axes of X and Y, and a 
variable ordinate, approaches the limit 2-n-a 2 , as the distance of the 
ordinate is indefinitely increased. 

J^dx 
— , we find 
x 

J" 6 dx 
— = Lim^^ log b — oo . 
i x 

Jr 00 dx 
— has no 
1 25 

meaning. 

242. Infinite Values of f(x). In the definition of I f(x)dx, f(x) 

is assumed to be a continuous function from x = a to x = b. If 
/(#) is continuous for all values from a to b except x = a, where it 
is infinite, the definite integral may be defined thus : 



fix) dx = Lim A=0 I / (x) dx. 



316 INTEGRAL CALCULUS 

If f(b) = oo, f(x) being continuous for other values of x, 

Xb S*b-h 

f(x) dx = Lim^ I f(x) dx. 

For example, consider Ex. 9, following Art. 239, 



r 



dy 



2 Va 2 -/ 



Here — , = oo , when y = a. 

Va 2 — y 2 

Hence 

J* ^/a 2 —y 2 J l -Va 2 -y 2 V a 6J 

_ 7T 7T IT 

Another example is Ex. 13, following Art. 239. 

r * 



■y/(x-2)(3-x) 
V(a>-2)(3-a;) 



Here — - = oo, when x = 2, and also when # = 3. 



Hence j — = Lim ft=0 j — 

J2 ■yj(x_2)(3-x) j2 + h V(s 

Lim^o sin -1 (2 x — 5) 



'"*= Lim^Csin- 1 (1-2 ft)- sin- 1 (-l + 2 ft)] 
= sin -1 l — sin _1 (— 1)= w. 

If /(a?) is infinite for some value c between a and 6, and is con- 
tinuous for other values, the definite integral should be separated 
into two. 

Cf(x)dx = Cf(x)dx+ Cf(x)dx. See Art. 243. 
These new definite integrals may be treated as already explained. 



INTEGRATION AS A SUMMATION 317 

EXAMPLES 

3 r - dx =l0 g(2+V3). 4. C ^-— = -log2. 

243. Change of Limits. The sign of a definite integral is changed 
by the transposition of the limits, 

f a f(?) dx = -£ h f(fi) dx. 

This is evident from (1), Art. 238, and also from the definition. 
For if x varies from b to a, the sign of Ax is opposite to that where x 
varies from a to b. Hence the signs of all the terms of ]x f(x) Ax 
will be changed, if the limits a and b are transposed. 

f(x)dx = — I f(x) dx. 

A definite integral may be separated into two or more definite 
integrals by the relation, 

f 7(a) dx = f /(*) dx + f /(a) (to. 

This follows directly from the definition. 

244. Change of Limits for a Change of Variable. When a new 
variable is used in obtaining the indefinite integral, we may avoid 
retaining to the original variable, by changing the limits to corre- 
spond with the new variable. 

For example, to evaluate 



p_* L _ i 

Jo 1 -f- V# 



assume y/x = z. 



318 

Then we have 



INTEGKAL CALCULUS 

dx 2zdz 



1+Vz 1 + z 
Now when x = 4, z — 2 ; and when x = 0, 2! = 0. 

Hence f —^p = C ^ « 2 [* - l og (1 + *)] 



= 4 - 2 log 3. 



EXAMPLES 



1. CxVx~T2dx = ^. Let a-f2 = z 3 . 
J2 15 

2. f( a; -2)" !K c« a; = - 3m + 5 
*/2 n 2 



+ 3^ + 2 



Let x — 2 = 2. 



3 f 2 s 8 ^ 3 (3 + ^g). Let a 2 + l=z 3 . 

J o(x 2 -\-iy 8 






2 ax — x 2 dx = 



7rcr 



Let x — a = a sin 6. 



5. r v3 + 2^-^ &=V3 _T Let a . =sl+ 2sinft 

c/2 (a; — l) 2 3 

6- I V(x — a)(b —x)dx = ^ (b — a) 2 . Let # = a cos 2 <£ + 6 sin 2 <£. 
7 j^V-^) f ^ = - 



3^ 

32 



Let x = a sin 3 0. 



8 \ x^a 2 -x 2 dx^(^ + ^\a\ Let z = asin0 



V64 48 



INTEGRATION AS A SUMMATION 319 

245. Definite Integral as a Sum. In the application of integration 
it is often convenient, in forming the definite integral from the data 

/(as) dx as the sum of an infinite number 

of infinitely small terms, f{x)dx being called an element of the 
required definite integral. 
From this point of view, 

ff(x) dx =f(a) dx + f(a. + dx) dx+f(a + 2dx)dx+ ••• +/(&) dx. 

This may be regarded as an abbreviation of the definition of a 
definite integral given in Art. 238. 



CHAPTER XXVIII 

APPLICATION OF INTEGRATION TO PLANE CURVES. 
APPLICATION TO CERTAIN VOLUMES 

246. Areas of Curves. Rectangular Coordinates. We have already- 
used this problem as an illustration of a definite integral. We will 
now consider it more generally, and derive the formula for the area 
in rectangular coordinates. 

247. To find the area between a given curve, the axis of X, and two 
given ordinates AP and BQ ; that is, to find the area generated by the 
ordinate moving from AP to 
BQ. 

Let OA = a, OB = b. 

Let x and y be the coordi- 
nates of any point P 2 of the 
curve ; then 

x + Ax, y + Ay, 

will be the coordinates of P 3 . 

The area of the rectangle 
P 2 A 2 A 3 is 

P 2 A 2 x A 2 A Z = y Ax* 

The sum of all the rectangles PAA U P X A^A 2 , P 2 A 2 A 3 , •••, maybe 
represented by 2\ a y &v* 

The required area PQBA is the limit of the sum of the rectangles, 
as A# is indefinitely diminished. That is 




= J y dx, 



* By Art. 245, one readily sees that this rectangle is an element of area. 

320 



APPLICATION OF INTEGRATION TO PLANE CURVES 321 



the lower limit a = OA, being the initial value of x, and the upper 
limit b = OB, the final value of x. 

Similarly the area between the curve, the axis of Y, and two 
given abscissas, GP and IIQ, is 



=J vdij, 



the limits of integration being the initial and final values of y, 
g=OG, and h= OH. 

EXAMPLES 

1. Find the area between the parabola y 2 = kax and the axis of 
X, from the origin to the ordinate at the point (h, k). 



Here A = f ydx= f h 2a^dx 



4a^i* 4a*7i- 



Since k 2 =lah, k = 2 aw, which 
gives 

A = hi2ah* = -hk = -OMPN* 




2. Find the area of the ellipse 

* + t = l. 

a 2 ^b 2 



Area BOA 



| y dx = - I Va- — x- (he 
o aJo 

bV x /-= t, . a 2 . _iX~\ a 

= -\ -V« 2 -ar + -sin '- 
a\_2 2 aj 



Tr«b 




* In finding areas, after the element of area and the limits of integration are 
chosen, the problem becomes purely mechanical. 



322 



INTEGRAL CALCULUS 



The entire area = -n-ab. 

Or we may integrate by letting x = a sin <£. 

Then f a Va*-a?dx = a* C\m 2 4>d<t>=^ C (1 -cos 2 <l>)d<f>= — 



Areai^^.^^ 
a 4 4 



3. Find the area included between the parabola x 2 = 4 ay, and the 



witch y = 



8<r 



# 2 + 4 d 2 



^Ins. (2,r-*V. 



Having found the point of intersection P, (2 a, a), we proceed as 
follows : 




Area AOP= AOMP- OMP* 

= C 2a $a?dx r 2a a?dx = 2 2 a 2 
Jo cc 2 + 4a 2 Jo 4 a 3 



Area between two curves = ( 2 v — )a 2 . 



4. Find the area of the parabola 

(y-5)* = 8(2-x), 
on the right of the axis of Y. 



Arts. 104. 



Length of element of area is the y of the witch minus the y of the parabola. 



APPLICATION OF INTEGRATION TO PLANE CURVES 323 

5. Show that the area of a sector of the equilateral hyperbola 
x 2 — y 2 = a* included between the axis of X and a diameter through 

the point (x f y) of the curve, is — log i ' • 



6. Find the entire area within the curve (Art. 133) f-Y + f£\ = 1. 

Ans. - wab. 
4 



7. Find the entire area within the hypocycloid (Art. 132) 

8 



ft + y$= a\ Let x = a sin 3 <£. ^?is. 3 



8. Find the entire area between the cissoid (Art. 125) y 2 = 



2a — x 
and the line x = 2 a, its asymptote. Arts. 3 ?ra 2 . 



9. Find the area of one loop of the curve (Art. 133) a*y 2 = a 2 x 4 — jc 6 , 

TO 

8 



Ans. ??. 



Also from x = - to a; = a. 



^3 8 ; 4 



10. Find the area of the evolute of the ellipse (Art. 167) 

(*r)^ + (6y)*= (a 2 - b 2 )K Ans. ?ffe - * 

8 \b a 

11. What is the ratio between a and 6, when the areas of the 
ellipse and its evolute are equal ? 

6 V3 

12. Find the area included between the parabolas 
y 2 = ox and x* = by. Ans. — • 



324 INTEGRAL CALCULUS 

13. Find the area included between the parabola 

y 2 = 2 x and the circle y 2 = 4 x — x 2 . Ans. 0.475. 

14. Find the area included between the parabola 

y 2 == 4 ax and its evolute (Art. 167) 27 ay 2 = 4 (x — 2 a) 3 . 

Ans . 352V?„ 2 . 
15 

Parametric Equations. Instead of a single equation between x and y 
for the equation of a curve, the relation between x and y may be ex- 
pressed by means of a third variable. Thus the equations 

x = a sin <£, y = a cos <j>, (1) 

represent a circle; for if we eliminate <£ from (1) we have 
x 2 + y 2 = a 2 (sin 2 </> + cos 2 <£) = a 2 . 

Equations (1) are called the parametric equations of the circle, and 
the third variable <f> is called the parameter. 

The formula A = I y clx is applied to (1) by substituting 
y = a cos <}>, dx = a cos </> d<j>. 

For a quadrant of the circle 

n 

A= I ydx= I a 2 cos 2 cf> d$ = 



7T(T 

T 



15. Find the area of one arch of the cycloid 

x = a (0 — sin 6), y = a (1 — cos 6). Ans. 3 ira 2 . 

16. The parametric equations of the trochoid, described by a point 
at distance & from the centre of a circle, radius a, which rolls upon a 
straight line, are 

x = a6 — b sin 0, y = a — b cos 0. 

Find the area of one arch of the trochoid above the tangent at the 

lowest points of the curve. 

Ans. it (2 a + b) b, when b < a or b > a. 



APPLICATION OF INTEGRATION TO PLANE CURVES 325 

248. Areas of Curves. Polar Coordinates. To find the Area POQ 
included between a Given Curve PQ and Two Given Radii Vectores ; that 
is, to find the area generated by the 
radius vector turning from OP to 
OQ. 

Let POX=a, QOX = f3. 

Let r and be the coordinates of 
any point P 2 of the curve, then 

r + Ar, + A0, 

will be the coordinates of P 3 . 

The area of the circular sector 
P 2 OR, is' 

- OPo x P.Mo = -r -r A0 = Ir 2 A<9. 
2 " " " 2 2 

The sum of the sectors POP, PxOR^ P 2 OR 2 , • ••, may be rep- 
resented by s -i 

The required area POQ is the limit of the sum of the sectors, 
as A0 approaches zero. That is, 




-IX'"* 



the initial value of 6, a = POX, being the lower limit, and the 
final value of 6, j3= QOX, the upper limit. 



EXAMPLES 

1. Find the area of one loop of the curve (Art. 144) r = a sin 2 0. 

A = lC 2 ,*dO = l fa 2 sin 2 2 0d$ = f f\l - cos 4 0)d0 

-V/Q J»/o 4 */0 

-IT- ' - 



7rd 2 



326 INTEGRAL CALCULUS 

The entire area of the four loops 
which is half the area of the circumscribed circle. 

2. Find the entire area of the circle (Art. 135) r = a sin 0. 

Ans. — — . 
4 

In the two following curves find the area described by the radius 
vector in moving from = to =-. 

3. r = sec0 + tan0. Ans. 1 + V2 - ^ 

8 

4. r = a(l-tan 2 0). Ans. fc-*\<A 

5. Find the entire area of the cardioid (Art. 141) r = a(l — cos 0). 

Ans. "* , or six times the area of the generating circle. 

Also find the area from = - to = — . Ans. (Sir — 2) — 

4 4 v ; 8. 

6. Find the area described by the radius vector in the parabola 

r = a sec 2 -, from = to $ = -. Ans. — 

Also find the area from = 5 to = — ^. ^4ns. — — 

3 3 9V3* 

7. Find the entire area of the lemniscate (Art. 143) r 2 = a 2 cos 2 0. 

^.fts. a 2 . 

8. Show that the area bounded by any two radii vectores of the 
reciprocal spiral (Art. 137) r0 = a is proportional to the difference 
between the lengths of these radii. 

9. In the spiral of Archimedes (Art. 136), r = a$, find the area 
described by the radius vector in one entire revolution from = 0. 

Ans. ^ 2 



2 
Also find the area of the strip added by the nth revolution. 

Ans. 8 (w — 1) ir 3 a 2 . 






APPLICATION OF INTEGRATION TO PLANE CURVES 327 



10. Find the area of the part of the circle 

/■ = a sin + b cos 6, from = to 6 = -. 



11. Find the area common to the two circles 

r = a sin + b cos 0, r = a cos + b sin 0. 

_, b\ a 2 + b 2 a 2 - b 2 



Ans. «•(«'+*$ + «» 

8 2 



^iws. 



+ 2 tan- 



where a > 6. 



12. Find the area of the loop of the Folium of Descartes (Art. 127) 



r = 



3 a tan sec 
1 + tan 3 



Ans. 



3 a 2 



13. Show that the line r= 2asec0 ( x + y=2a), divides the 

l + tanfl' V J J 

area of the loop of the preceding example in the ratio 2:1. 

a 

14. Find the entire area within the curve (Art. 145) r = a sin 3 -, no 

o 



part being counted twice. 



Ans. (10tt + 9V3) §-. 



249. Lengths of Curves. Rectangular Coordinates. To find the 
Length of the Arc PQ between Two Given Points P and Q. 
Let OA = a, OB = b. 

Denoting the required length of 
arc by s, we have from (1), Art. 155, 



ds 



=nMS 



\dx 



dx. 






Hence 



wi + aw«. 



and between the given limits 



s>m 



a P. 



dx, • • (1) A 

the limits being the initial and final values of x. 



B X 



328 INTEGRAL CALCULUS 

We may also use the formula 



-&+&« « 



the limits being the initial and final values of y, 
g=OG, and h=OH. 



EXAMPLES 



1. Find the length of the arc of the parabola y 2 = 4=ax, from the 
vertex to the extremity of the latus rectum. 



Here 4 = 2!. 

therefore $ =j[Yl + ~Y <& = f Y£±*V <fo,. 
This may be integrated by Ex. 13, p. 305, making 6 = 0. 
rfa±x\l dx = ^ ax + x 2 + a log ( V"^+^ + Vas) 

j^Va + »y ^ = o j-^2 + log £ + ^2)] =2.29558 a. 
Or we may use the formula (2), 



*=jhr 



_ ?/ 2 dx _ y 
4 a cfa/ 2 a 



= M| V^HU 2 + ^ log (2/ + V2/ 2 + 4a 2 )T 
= a[V2 + log(l+ V2)] 



APPLICATION OF INTEGRATION TO PLANE CURVES 329 

2. Find the length of the arc of the semicubical parabola 
(Art. 130) ay 2 = x 3 , from x = - to x = 5a. Ans. — . 

3. Find the entire length of the arc of the hypocycloid (Art. 132) 

2. 2. 2 

x*+y 3 = a 3 ' Ans. 6 a. 

4. Find the length of the arc of the catenary (Art. 128) 

a - — 
y = -( e a + e -), 

^ a - - x 

from x = to the point (x, y). Ans. ^{e a — e °). 

5. Find the length of the arc of the curve 

y = log sec x, from x = to x = -. 

Ans. log (2 + V3). 

6. Find the length of the curve 

17 

6 aw = x* + 3, from x = 1 to a: = 2. ^4ns. — . 

* ' 12 

7. Find the perimeter of the loop of the curve 

9 a y- =(x-2 a) (x - 5 a) 2 . Ans. 4 V3 a. 

8. Find the length of that part of the evolute of the parabola 
(Art. 167) 27 ay 2 = 4 (x — 2 a) 3 included within the parabola y 2 = 4 ax. 

Ans. 4(3V3-l)a. 

9. Find the length of the curve 

y = log — = — , from # = 1 to x = 2. 

eZ + l Aiis. log(e + e _1 ). 

10. Find the length of one quadrant of the curve ( - j + ( - ) = 1. 

Ans. ^±^±&_ 2 . 
a + b 

11. The parametric equations of a curve are x = e 9 sin 0, y = e e cos 0. 
Find the length of arc from = to 6 = jj. -4n*. V2 (e*- 1). 

40 



330 



INTEGRAL CALCULUS 



12. The parametric equations of the epicycloid, the radius of the 
fixed circle being a, and that of the rolling circle ~, are 

z 



x = - (3 cos 4> — cos 3 $), 



y — - (3 sin <£ — sin 3 <£), 



<f} being the angle of the fixed circle, over which the small circle has 
rolled. 

Find the entire length of the curve. Ans. 12 a. 

250. Lengths of Curves. Polar Coordinates. To find the Length 
of the Arc PQ between Two Given Points P and Q. 
Let POX = a, QOX^fi. 
We have from (3), Art. 156, 



ds 


= p 2 + 


(1)7- 


rei 
s 


ore 


2 fdrVJA 

\dB) _ 



». • • (1) 



the limits being the limiting values q' 
of 0. 




Or we have ds 
therefore s 



1 + r 



Off 






dr 
-f-r 2 



dr 



(2), Art, 156, 



dr 



dr, 



(2) 



the limits being the limiting values of r. That is, OP— a, OQ = b. 



EXAMPLES 



1. Find the length of the arc of the spiral of Archimedes 
(Art. 136), r = aO, from the origin to the end of the first revolution. 



Here 



— = a, and we have by (1), 
d6 



APPLICATION OF INTEGRATION TO PLANE CURVES 331 

dO 



s= C~\a 2 6 2 + a 2 )?d0 = a C*" (1+0 2 )"' 



=0 [^l±i: + |i og( , + vr+^)J 



TtVI + 47T 2 +hog (2 7T + Vl + 47T 2 )"]. 



Or ^ve ma}' use the formula (2) 



Jf \i 



'06 



a r dO 1 

a ar a 



s = 



Jf»2jra 
Vl + r 2 h dr = 1 p " V?"^ 2 dr 
\ a 2 a «/o 

= * [~| V^+d 2 + flog (r + Vt : M^)1 2 

= a 7rV47r 2 + l+^log(27r + vW + l)l. 



2. Find the entire length of the circle (Art. 135) r = 2 a sin 0. 

Ans. 2 ira. 

3. Find the length of the arc of the circle 



r = a sin + & cos 0, from = to (r, 0). ^Lns. Va 2 + b 2 0. 

4. Find the length of the logarithmic spiral (Art. 138) r = e ae from 
the point (r 1} 0j) to (r 2 , 2 ), using the formula (2), and the equation 



O^Ml. Ans. Va2 + 1 (r 2 - ri ). 

a a 



332 INTEGRAL CALCULUS 

5. Find the entire length of the cardioid (Art. 141) 

r =a (1 — cos 0). Ans. 8 a. 

Also show that the arc of the upper half of the curve is bisected by 



6. Solve Ex. 5 by using formula (2) and the equation 6 = vers -1 -. 

7. Find the arc of the reciprocal spiral (Art. 137) r6 = a, from 
$ = £ to 6 = ?. Ans. (~ + log-V 



12 4 V15 



8. Find the arc of the parabola (Art. 139) r — a sec 2 - from 

$ = to 6 = |. ^4ns. (sec ^ + log tan *Lz)a. 



9. Find the entire length of the arc of the curve (Art. 145) 

r = a sin 3 1 ^. ?£« 

Also show that the arc AB is one third of OABG. 
Hence OA, AB } BG are in arithmetical progression. 

10. Find the entire length of the curve r = a sin n -, n being a posi- 
tive integer. 



See for integration Exs. 24, 25, p. 314. 

2.4-6... 



Ans. 2 a, when n is even 



1.3.5...(?i-l) 

1.3-5-w 
2.4-6... (n-1) 



7ra, when n is odd. 



APPLICATION OF INTEGRATION TO PLANE CURVES 333 

251. Volumes of Surfaces of Revolution. To find the Volume gener- 
ated by revolving about OX the Plane Area APQB. 

Let OA = a, OB = b. 

Let x and y be the coordinates 
of any point P 2 of the given 
curve. 

It is evident that the rectan- 
gle P,A. 2 A~ will generate a right 
cylinder, whose volume is 

TTlf A.i\ 

The sum of all these cylinders 
may be represented by 




Ai A 2 A 3 A 4 B 



The required volume is the limit of the sum of the cylinders, as 
Ax approaches zero. That is, 

Similarly the volume generated by revolving PGHQ about OF is 

V y = ir\ x 2 cbj, 
where OG = g, and OH = It. 



EXAMPLES 
1. Find the volume generated by revolving the ellipse 

9 9 

a 2 + Z> 2 ' 
about its major axis, OX. This is called the prolate spheroid. 



— =ttI y 2 dx = 7r I — (a 2 — a?)dx = —- 



crx 



arY 27ra^ 2 



F=%a6 2 . 



334 



INTEGRAL CALCULUS 



2. Find the volume generated by revolving the ellipse about its 
minor axis, OY. This is called the oblate spheroid. 



|_j^=^j>-^= 



2 7ra?b 



y=* 7T a 2 b. 
3 



3. If the parabola y 2 = kax is revolved about OX, show that the 
volume from a? = to x=2a is one third the volume from x = 2 a 
to x = 4 a. 



4. Find the volume generated by revolving the segment LOL' of 
the parabola about the latus rectum LL'. 



Here -?= tt C\pN) 2 dy = ir C\a - x) 2 dy 
C "-"( y- V r , 16 



7rCT 



Ans. 



32 
15 



irCT. 



5. Find the volume generated by revolving 
about OX one loop of the curve (Art. 134) 



a A y 2 — a 2 x A — x e 



Ans. — -n-a 3 . 
35 



Y 


P/ 


t 

N 
2a 





a 


v 


F X 
L' 



6. Find the entire volume generated by revolving about OX the 



hypocycloid (Art. 132) x*+y* = a*. 



Ans. 



32 
105 



7r(r. 



7. Find the volumes generated by revolving about OX, and 



about OY, the curve (Art. 133) 



Ans. V x = ^irab 2 . V y = ^a 2 b. 
oo o 



APPLICATION OF INTEGRATION TO PLANE CURVES 335 

8. The part of the line - + | = 1, intercepted between the coor- 

a o 

dinate axes, is revolved about the line x = 2 a. Find the included 
volume. Ans. -vefb,, 

9. The segment of the parabola, x 2 — 3 x + 2 y = 0, above OX, is 

81 
revolved about OX. Find the volume generated. Ans. — -. 

10. A segment of a circle is revolved about a diameter parallel to 
its chord. Show that the volume generated is equal to that of a 
sphere whose diameter is equal to the chord. 

11. Find the volume generated by revolving about OF the witch 

(Art. 126), y = 8a ' „ , from (0, 2 a) to y = a. Ans. 4 (log 4-1) 7ra 3 . 
or + 4 a 2 

12. Find the volume generated by revolving the upper half, ABA', 
of the curve (Art. 133) [- ] + ( -) =1, about the tangent at B. 

Ans. ^-ffW 

4 3oJ 

13. Find the volume generated by revolving about OX the area 

included between the ellipse — -|- ^ = 1, and the parabola 2 ay 2 = 3 b 2 x. 

Ans. — irab 2 . 
48 

X X 

14. A segment of the catenary (Art. 128), y = ~{e a -\-e a ), by a 

chord through the points x = ± a log 2, is revolved about the tangent 
at the vertex. Find the volume generated. 

Ans. 3nog2-^J7ra 3 . 

15. Find the volume generated by revolving about the latus rec- 
tum of the ellipse — + ^ = 1, the segment cut off by the latus 
rectum. Ans. 2ir{ab 2 -—- ab^/d 2 - b 2 sin" 1 -\ 



336 



INTEGRAL CALCULUS 



252. Derivative of Area of Surface of Revolution. In order to 
obtain the formula for the surface generated by the revolution of a 
given arc, it is necessary to find the derivative of this surface with 
respect to the arc. 

Let S denote the surface gen- 
erated by revolving about OX 
the arc s, AP. 

Using for abbreviation the 
expression "Surf ( )" to denote 
"the surface generated by re- 
volving ( ) about OX" we have 

S = Surf (s), AS = Surf (As). 



This may be written 




AS 



Surf (As) 



Surf (Chord PQ) 



Surf (Chord PQ) (1) 



Now the surface generated by the chord PQ is the convex surface 
of the frustum of a right cone, which is the product of the slant 
height by the circumference of a section midway between the bases. 

Hence Surf (Chord PQ) = 2 tt (?E±M\ chord PQ 

= 2 ^±1^ Chord PQ 

= 7r (2y + Ay) Chord PQ. 

Substituting this for the last factor in (1), and dividing both sides 
by As, we have 

Ag = Surf (A,) , (2 +A) Chord PQ 
As Surf (Chord PQ) ^ 9 ^ 9 > As 



Taking the limit of each member, as As approaches zero, noticing 
that 



Lim 



As=0 



Surf (As) 



Surf (Chord PQ) 



1, 



APPLICATION OF INTEGRATION TO PLANE CURVES 337 



and 






Lim A „ => ChOTdP( ? = l, we ha* 



f = Lim irf 
ds 



As 

AS 



As 



Try. 



Similarly if Y is the axis of revolution, 
dS 



ds 



2ttx. 



253. Areas of Surfaces of Revolution. To find the Area of the Sur- 
face generated by revolving about OX the Arc PQ. 

By the preceding article we have 

dS 



ds 



= 27ry 



hence S = I 2-rry ds. 

To express this in terms of x and 
y, we have from (1), Art. 155, 



ds 



MSI 



dx, 



>n* 



which gives 

*— £{HWJ 

If Y is the axis of revolution, 



dx. . (1) 




dyYf 
dx) 



dx. 



S y = 2 7T Cxds = 2 7T C x\l + 

Or we may use rls = \ 1 + ( -^ ) dy, 

and instead of (1) we have 

and instead of (2) S v = 2ir Cx[l + (— Y~] dy. 



B X 



(2) 



(20 



338 



INTEGRAL CALCULUS 



EXAMPLES 

1. Find the area of the surface generated by revolving about OX 
the hypocycloid (Art. 132) o$ + y* = a*. 



Here 



(y = J - a,i)4, ^ = - (a* -x^x~\ 
ctx 



Using (1) ±S x = 2ir£(a} 



£C 3 ) : 



1 + 



2. 2._,1 



a 3 — x s ~\~ 
x* J 



2 •.{ /Y3 j /^ a 2 2 3 1 

x*y—dx = 2 7ra 3 | (a 3 — # 3 ) * a; - 3 cto 



i«3 



6 7ra 2 



#* = 



12 W 



Or we may use (1') -^ = — (a 3 — 2/*) 2 2/" 



■& = 2 



-X' 



1 + 



2 :2_1 

a 3 — y 

2 
F J 



1 f* a 1 

dy = 2 ?ra 3 I y^dy = 



6-n-a 2 



2. Show that the area of the surface generated by revolving the 
parabola y 2 = 4 ax, about OX, from a? = to x = 3 a, is one eighth of 
that from x — 3atox = 15a. 



3. Find the area of the surface generated by revolving about OX 
the loop of the curve 9 ay 2 = x(3a— xf. Ans. 3 tto 2 . 

4. Find the surface generated by revolving about OX, the arc 
of the curve 6 a 2 xy = $ 4 + 3 a 4 , from x = a to x = 2 a. *„ 

Ans. -—tto 2 . 
16 

5. The arc of the preceding curve from x = a to x = 3 a, revolves 
about Y. What is the surface generated ? Ans. (20 + log 3)?ra 2 . 

6. Find the surface generated by revolving about OF the curve 
4 y = x 2 — 2 log x, from x = 1 to x = 4. Ans. 24 tt, 



APPLICATION OF INTEGRATION TO PLANE CURVES 339 

7. Find the entire surface generated by revolving about OX the 
ellipse 3.r + 4?/- = 3cr. Ans (8 + _*\ rC p 

\ 2 VsJ 

8. Find the entire surface generated by revolving about OY the 
preceding ellipse. Am (4 + 3 i og 3)i£ 

9. Find the surface generated by revolving about OX the loop 

of the curve S aV = a-x 2 — x*. A ira? 

J Ans. 

4 

10. An arc. subtending an angle 2 a, of a circle whose radius is a, 
revolves about its chord. Find the surface generated. 

Ans. 4 7ra 2 (sina — a cos a). 

ci f - — -\ 

11. The arc of the catenary (Art. 121) y = -±le« + e « ), from x = a 



to x=2a, revolves about OT. Find the surface generated. 

Ans. (e> + 2e- l -3e- z )ira 2 . 

12. The parametric equations of a curve are 

x = e e sin 6, y = e e cos 0. . 

Find the surface generated by revolving the arc from 6 = to 

0=-, about OX Ans. i*(e»-2). 

2 5 

13. Find the surface generated by revolving about OT the arc of 

the preceding example. AnSm ^ 7r (2c 77 1 1). 

o 

14. The parametric equations of the epicycloid, the radius of the 
fixed circle being a, and that of the rolling circle - (Ex. 12, p. 330) 

are x = — cos <f> cos 3$, y = — sin <£ — - sm 3 <£. 

Find the entire surface generated by revolving the curve about OX. 

Ans. - 7r 2 a 2 . 

15. Find the surface generated by revolving one arch of the pre- 
ceding curve about OY. Ans. drra 2 . 



340 



INTEGRAL CALCULUS 



254. Volume by Area of Section. The volume of a solid may be 
found by a single integration, when the area of a section can be ex- 
pressed in terms of its per- 
pendicular distance from 
a fixed point. 

Let us denote this dis- 
tance by x, and the area 
of the section, supposed to 
be a function of x, by X. 

The volume included 
between two sections sep- 
arated by the distance dx 
will ultimately be Xdx, 
and we have for the volume of the solid 

V= Cxdx, 

the limits being the initial, and final, values of x. 




EXAM PLES 

1. Find the volume of a pyramid or cone having any base. 

Let A be the area of the base, and h the altitude. 

Let x denote the perpendicular distance from the vertex of a sec- 



tion parallel to the base 
have, by solid geometry, 

X = tf 
A h 2 ' 

Hence, 



Calling the area of this section X, we 



X = 



Xdx = - ) I 

/iVO 



x 2 dx 



Ax 2 
Ji 2 ' 



Ah s 
h 2 3 



Ah 

3 ' 



2. Find the volume of a right conoid 
with circular base, the radius of base being 



a, and altitude h. 




OA = BO=2 a, BO=CA = h. 
The section BTQ, perpendicular to OA, is an isosceles triangle. 



APPLICATION OF INTEGRATION TO PLANE CURVES 341 
Let x = OP; then 



X = area PTQ = PTxPQ = h V2 ax - x\ 

Hence, V= f* Xdx = h C**^/2ax- x 1 dx = — ■ 
«/o «/o 2 



This is one half the cylinder of the same base and altitude. 



3. Find the volume of the ellipsoid 



Let us find the 


* + £ + $=!• 

(V 0~ C" 


7 


.... a 


area of a section 








C'B'D' perpendicular 
to OX, at the dis- 


^^^f C 


c 

\ 
\ 


>£i 


tance from the origin 


/ T — *~7 


"1sk\ 


031= x. 


x a 




y n 


This section is an 






ellipse whose semi- 
axes are MB' and 


^y 


i — * 




\ / 




MC 


y/ 




To find MB', let 




z = in (1), aud we 


7i 






have 


y = MB' = -^d 2 - 
a 


XT. 





To find J/C", let y = in (1), and we have 



z = MC = -->/a 2 -x*. 

a 



The area of the ellipse (Ex. 2, p. 321) is it (MB 1 ) (MC). 



Hence, 



— («--.i~), 



and V= 2 P X da; = " " ^ P (a 2 - s 2 ) da; = Kabc. 

«A> a- J 3 



342 INTEGRAL CALCULUS 

4. A rectangle moves from a fixed point, one side varying as the 
distance from the point, and the other as the square of this distance. 
At the distance of 2 feet the rectangle becomes a square of 3 feet. 
What is the vohime then generated ? Ans. 41 cubic feet. 

5. The axes of two right circular cylinders having equal bases, 

radius a, intersect at right angles. Find the volume common to 

the two. , 16 a 3 

Ans. 

3 

6. A torus is generated by a circle, radius b, revolving about an 
axis in its plane, a being the distance of the centre of the circle from 
the axis. Find the volume by means of sections perpendicular to 
the axis. Ans. 2 7r 2 a 2 b. 

7. A football is 16 inches long, and a plane section containing a 
seam of the cover is an ellipse 8 inches broad. Find the volume of 
the ball, assuming that the leather is so stiff that every plane cross- 
section is a square. Ans. 341^ cu. in. 

8. Given a right cylinder, altitude h, and radius of base a. 
Through a diameter of the upper base two planes are passed, touch- 
ing the lower base on opposite sides. Find the volume included 
between the planes. ^ / _ |\ % _ 

9. Two cylinders of equal altitude h have a circle of radius a, 
for their common upper base. Their lower bases are tangent to 
each other. Find the volume common to the two cylinders. 

Ans. i^. 



CHAPTER XXIX 
SUCCESSIVE INTEGRATION 

255. Definite Double Integral. — A double integral is the integral 
of an integral. 

Thus, x and y being independent variables, the definite double 
integral, „ a „ c 

I f(x,y)dxdy } 

indicates the following operations: 

Treating x as a constant, integrate f(x. y) with respect to y 
between the limits d and c ; then integrate the result with respect 
to sb between the limits b and a.* 

For example, 

J &(P - y)dx dy = J x-l by - 1 j dx = J x~~dx 

Xotice that the order of the two integrations is indicated in the 
given definite integral by the order of the differentials dxdy, taken 

| being nsed in the same 
order. Jo 

It should be said, however, that the order of the integrations is 
denoted differently by different writers. 

256. Variable Limits. — The limits of the first integration, instead 
of being constants, are often functions of the variable of the second 
integration. 

* Using parentheses, this might be repmsfntert by I [( f(z,y)dy\<l.r. 

343 



344 INTEGRAL CALCULUS 

For example, 
JP p xy dy dx = jT°g 2 J" j dy = | jf "(3 jf + 2 V - a*y)dy = ^ . 

As another example, 
J J^^(x-\-y)dxdy=J f^ + |-) o ^"^dx 






When the limits are all constants, as in Art. 248, the order of the 
integrations may be reversed without affecting the result. That is, 

i I x 2 (b — y)dxdy — I I x 2 (b — y)dy dx. 

a *S0 *^ */a 

Where the definite integral has variable limits, the order of integra- 
tions can be changed only by new limits adapted to the new order. 

257. Triple Integrals. — A similar notation is used for three suc- 
cessive integrations. Thus 

III x 2 y 2 zdxdydz = I I ——x 2 y 2 dxdy 

EXAMPLES 

Evaluate the following definite integrals : 

n 6 a 2 b 2 

®V 0» - V) dx dy = — (a - b). 

2. C Cr 2 sin 6 dr dB = a3 ~ 63 (cos /3 - cos a). 



SUCCESSIVE INTEGRATION 345 

Jb Jo r dr d0 = — • 

2 --± 

5. I r'sin^?^)-^. 

Jo Jo 3 

nlO y 
Vxy — y 2 dy dx = 6 a 3 . 

«/0 «A) 2 



8. fT' r^ear = L- 1 ±\^. 

c/o c/acose ^ I5y 10 

9- J^jTsin (2 +-«)<»** = 1. 

10. fTfsirf **«** = £+5. 

»/o »/o 6 8 

Jo J^ 2 2? + ^ 2 & 

a 

12 - X B XX ( ^ + y2 + * 2) dx dy dz = t (a2 + 52 + c2) - 



13. j j sin (xyz) dx dydz = f 



14. C f f v uvw du dvdw= — 
J" %/o Ju-% 18 

15 ' J S*f* VeX+y+zdxd y dz = - 



CHAPTER XXX 

APPLICATIONS OF DOUBLE INTEGRATION 

258. Moment of Inertia. If r 1} r 2 , ?* 3 , • ••, r n are the distances from 
a given line of n particles of masses m 1} m 2) m 3 , ■ • •, m n , the sum 

m^ 2 + m 2 r 2 + ?%r 3 2 + • • • -f m n r n 2 = X (w?* 2 ) 

is defined in treatises on mechanics as the moment of inertia of the 
system about the given line. 

The moment of inertia of a continuous solid about a given line is 
the sum of the products obtained by multiplying the mass of each 
infinitesimal portion of the solid by the square of its distance from 
the given line. 

The summation is then effected by integration, and we have for 
the moment of inertia of a body of mass M, 



-/- 



dM. 



259. Moment of Inertia of a Plane Area. The moment of inertia 
of a given plane area about a given point may be defined as the 
sum of the products obtained, by multiplying the area of each infini- 
tesimal portion by the square of its distance from 0. 

This may be regarded as the moment of inertia of a thin plane 
sheet of uniform thickness and density, about a line through per- 
pendicular to the plane, the mass of a square unit of the sheet being 
taken as unity. 

We shall illustrate double integration by finding the moment of 
inertia of certain areas. 

346 



APPLICATIONS OF DOUBLE INTEGRATION 



847 



M' N' 



i 



t 



T 



[Q 



— i. — i 



M N 



260. Double Integration. Rectangular Coordinates. To find the 
moment of inertia of the rectangle OACB about 0. 

Let OA = a. OB = b. 

Suppose the rectangle 
divided into rectangular 
elements by lines parallel 
to the coordinate axes. Let 
sb, y, which are to be re- 
garded as independent vari- 
ables, be the coordinates of 
any point of intersection as 
P, and x + dx, y + dy the 
coordinates of Q. Then the area of the element PQ is dxdy. 

Moment of inertia of PQ = OP' • dx dy = (x 2 + y 2 ) dx dy. 

The moment of inertia of the entire rectangle OACB is the sum of 
all the terms obtained from (x 2 + y 2 )dx dy, by varying x from to a, 
and y from to b. 

If we suppose x to be constant, while y varies from to b, we shall 
have the terms that constitute a vertical strip MNN'M'. 
Hence 

Moment of inertia of MNN'M' = dx I (x 2 + y 2 ) dy 



(. 



V 



dx[ x 2 ii + 
V 3yo 



M + ^yiae. 



Having thus found the moment of a vertical strip, we may sum all 
these strips by supposing x in this result to vary from to a. That 
is, the moment of inertia of OACB, 



1 = 



re + ?) 



,2 , 'A 7 , _ a?b + ab 3 



dx — 



The preceding operations are those represented by the double 
integral, 

T= C C b (pf + f)dxdy. 



348 



INTEGRAL CALCULUS 



If we first collect all the elements in a horizontal strip, and then 
sum these horizontal strips, we have 



1= f I (x 2 + y 2 ) dy dx = 



+ ab 3 




261. Variable Limits. To find the moment of inertia of the right 
triangle OAO about 0. 

Let OA = a, AC=b. The Y 
equation of 00 is 

b 
V = — x. 
3 a 

This differs from the pre- 
ceding problem only in the 
limits of the first integration. 

In collecting the elements in a vertical strip MN, y varies from to 
MN. But MNis no longer a constant as in Art. 260, but varies with 

OM. according to the equation of OC, y = -x. 

b a 

Hence the limits of y are and -x. 

a 

In collecting all the vertical strips by the second integration, x 
varies from to a, as in Art. 260. 

Thus we have for the moment of inertia of OAO, 



*/0 



+ y*)dxdy = ab(?- + 



12 



By supposing the triangle composed of horizontal strips as HK, 

we shall find 

Y 

I- 



'= I I (xP + y^dydx 

T 

■G + 5> 




APPLICATIONS OF DOUBLE INTEGRATION 349 

262. Plane Area as a Double Integral. If in Art. 260 we omit the 
factor (ar^ + y 2 ), we shall have, instead of the moment of inertia, the 
area of the given surface. 



That is, Area = } ) dx dy = I J dy dx, 



the limits being determined as before. 



EXAMPLES 

1. Find the moment of inertia about the origin of the right tri- 
angle formed by the coordinate axes and the line joining the points 

»),(0,6). b(a-*) 

2. Find the moment of inertia about the origin of the circle 
r +>/- = «-. 



Ans. i £ j^- x \x* + y^dxdy = ^. 



3. Find by a double integration the area between a straight line 
and a parabola, each of which joins the origin and the point (a, 6), 
the axis of X being the axis of the parabola. 



h\l x ^ 

Ans. I I dx dy = I I dy dx = — 



4. Find the moment of inertia about the origin of the preceding 

5. Find by a double integration the area included between the 
circle xr + y- = 10 ay, the line 3 x -j- y = 10 a, and the axis of T. 

10 q-y 
>~a /»Vl0ay-y2 /M0a •» 3 

Ans. dy dx + I dy dx 

/-Sa /*10a-3z g y . gy 

= I I , dx dy = — - 3 +5 sin x - 



350 



INTEGRAL CALCULUS 



6. Find the moment of inertia about the origin of the area be- 

tween the ellipse — f-*_ = l and the line - + " = 1. 
a- b- a b 

Arts. .[- 



(jL-A-\ a sb + ab 3 ). 
\16 12/ J 



7. Find the moment of inertia about the origin of the area be- 
tween the parabola ay = 2 (x 2 — a 2 ), the circle x 2 + y 2 = a 2 , and the 



axis of Y 



.35 



8. Find by a double integration the area included between the 
parabolas y 2 = 3x, and y 2 = 12 (60 — x). Ans. 960. 



9. Find the moment of inertia of the area included between the 
parabola y 2 = 4 ax, x = 4=a } and the axis of X, about the focus of the 
parabola. . A . 2336 



Ans. 



35 



a\ 



10. Find the moment of inertia of the area included between the 
lines y = 2x, x + 2y = 5a f and the axis of X, about the intersection 



of the first two lines. 



Ans. 



125 a 4 



\M' 



/ ;■ 



M 



263. Double Integration. Polar Coordinates. To find the area of 
the quadrant of a circle AOB, whose radius is a 

In rectangular coordinates, Art. 
260, the lines of division consist of 
two systems, for one of which x is b 
constant, and for the other, 2/ is l 
constant. L 

So in polar coordinates, we have 
one system of straight lines through 
the origin, for each of which is 
constant, and another system of 
circles about the origin as centre, for 
each of which r is constant. 

Let r, 0, which are to be regarded 
as independent variables, be the coordinates of any point of intersec- 






l/0c^lxA"\'"\ "' 



N N' A 



APPLICATIONS OF DOUBLE INTEGRATION 



351 



tion as P, and r + d)\ 4- dO, the coordinates of Q. Then the area 

of PQ is ultimately 

PBxPQ=rdO'dr. 

If we first integrate, regarding constant while r varies from 
to a. we collect all the elements in any sector MOM'. 

The second integration sums all the sectors, by varying 6 from 



*>l 



Hence Area BOA = f ~ C r dO dr = — 



If we reverse the order of integration, integrating first with respect 
to 6, and afterwards with respect to r, we collect all the elements in 
a circular strip XLL'X', and sum all these strips. This is written 



Area BOA = 



XT' 



dr dO. 



264. If the moment of inertia about is required, we have for 
the moment of inertia of PQ, r-rdOdr. Hence, the moment of 
BOA is 



1= C C r\Wdr= f f 2 r*drdO = 






265. Variable Limits. To find by a double integration the area 
of the semicircle OB A with radius OC=a, the pole being on the 
circumference. 

The polar equation of the circle is 
/* = 2 a cos 0. If we integrate first 
with respect to r, then with respect 
to 6, we shall have 



Area OB 



-IT 



dOdr = 



-<i- 




Here, in collecting the elements in a radial strip OM, r varies 
from to OM. But 03/ varies with 0, according to the equation 
of the circle r = 2 a cos 6. Hence the limits are and 2 a cos 6. 



352 



INTEGRAL CALCULUS 



In collecting all these radial strips for the second integration, $ 
varies from to - • 

By supposing the area composed of concentric circular strips 
about as LK, we find 




A X 



Area OBA 



X2a /»c 



'(£) 



r dr d6 = 



Tror 



EXAMPLES 

1. Find the moment of inertia about the origin of the area in- 
cluded between the two circles, r = a sin and r = b sin 0, where a > b. 

no sin G O 



sing 



32 



2. Find the moment of inertia about the origin of the area 
between the parabola (Art. 139), r = asec 2 -, its latus rectum, and 

0X ' Ans. i^- 4 . 

35 

3. Find the moment of inertia about its centre of the area of the 



lemniscate (Art. 143) r 2 = a 2 cos 2 0. 



Ans. 



irCC 



4. Find by double integration the entire area of the cardioid 
(Art. 141) r = a(l-cos0). 



Ans. 



Sva' 



5. Find the moment of inertia about the origin of the area of the 

preceding cardioid. A 35 wa 4 

Ans. -_- 



APPLICATIONS OF DOUBLE INTEGRATION 353 

6. Find the moment of inertia about its centre of the entire arc 

Df the four-leaved rose (Art. 144) r = a sin 2 0. A 3 wci* 

K J Ans. 

16 

7. Find by a double integration the area of one loop of the 

lemniscate (Art. 143) outside the circle 2r = a 2 . i / /« tt\ a 2 

Ans. (V3- 5 ) T 

8. Find the moment of inertia of the area of the preceding 
example about the centre of the lemniscate. * /V3 ir\ a 4 

RS ' V~2~ + 3)l6 

266. Volumes and Surfaces of Revolution. Polar Coordinates. If 
in the figure of Art. 263 we suppose a revolution about OX, the 
volume generated by the infinitesimal area PQ is the product of 
this area by the circumference through which it revolves, that is, 

r r sin • r dO dr. 

Hence for the entire volume 



V=2tt f frsmedBdr, 



the limits being determined as in Art. 263. 
If the revolution is about OY, 



V= 2 7T f Cr cos OdO dr. 
The area of the surface generated about OX is (Art. 253) 



U 



EXAMPLES 

1. Find the volume generated by revolving the cardioid (Art. 141) 

= a (1 — cos 6) about OX. . 8 , L . ^ ., , , 

Ans. - irflr, twice the inscribed sphere. 

2. Show that the entire volume generated by revolving the four- 

leaved rose (Art 144) r = a sin 2 6 about OX is — of the volume of the 
circumscribed sphere. 



354 INTEGRAL CALCULUS 

3. Find the volume generated by revolving one loop of the four- 
leaved rose r = a sin 2 about the axis of the loop. 

Ans. fs V2 - 9 V 



105 

4. Find the volume generated by revolving the lemniscate (Art, 

143) r 2 = a 2 cos 2 about Y. ^ a s vk 

-4ns. -^— . 



5. Find the volume generated by revolving the lemniscate about 
OX 



^4ns. 



L L log(V 2 +1 )-|; 



6. Find area of surface generated, by revolving the cardioid 
r=a(l -- cos 0) about OX. , 32 -n-a 2 



7. Find the moment of inertia of a sphere (radius a) about a 
diameter, m being the mass of a unit of volume. Ans. 8 7ra 5 m 



15 



CHAPTER XXXI 

SURFACE, VOLUME. AND MOMENT OF INERTIA OF ANY SOLID 

267. To find the Area of Any Surface, whose Equation is given 

between Three Rectangular Coordinates, x 3 y, z. 

Let this equation be „, 

z=f(x,y). 

Suppose the given surface to be divided into elements by two 
series of planes, parallel respectively to XZ and YZ. These planes 
will also divide the plane XY into elementary rectangles, one of 
which is PQ. the projection upon the plane XY of the corresponding 
element of the surface P'Q'. 




Let x, y, z be the coordinates of P' and x + dx, y + dy, z + dz, 
of (>'. 



356 



INTEGRAL CALCULUS 



Since PQ is the projection of P'Q', the area of PQ is equal to that 
of P'Q' multiplied by the cosine of the inclination of P'Q' to the 
plane AT. This angle is evidently that made by the tangent plane 
at P' with the plane XY Denoting this angle by y, 

Area PQ = Area P'Q' • cos y, 

Area P'Q' = Area PQ • sec y. 

We see from the figure that 

Area PQ = dx dy. 



Also from (8), Art. 110, sec y 



dz 



fdz 



dx J \dy 



where — and — are partial derivatives, taken from the equation of 
dx dy 

the given surface z =f(x, y). 

Hence Area P'Q' = [l + (^Y + (^ 



dx dy. 



If S denote the required surface, 



S 



-ff[ i+ 



© V: 



dz 
\dy 



dxdy, 



(i) 



the limits of the integration depending upon the projection, on the 
plane XY, of the surface required. 

For example, suppose the surface ABC to be one eighth of the 
surface of a sphere whose equation is 



Here 



1 + 



W 



X 2 


+ 2/ 2 + 


z 2 = a 2 . 




dz 

dx 


X 

— — > 

z 


dz __ 
dy 


z 


A2 


-1 , x 


l + f 


a 2 



a 2 — x 2 — y 2 



SURFACE. VOLUME, AXD MOMENT OF INERTIA 357 



Substituting in (1), we have 



dx dy 



J J Va 2 — s 



ir 



This is to be integrated over the region OB A, the projection of 
the required surface on the plane XY. 
The equation of the boundary AB is 



s» + 



a". 



Integrating first with respect to y, we collect all the elements in 
a strip M'X'KL. y varying from zero to ML, that is, between the 
limits and v<r — .r\ 

Integrating afterwards with respect to x, we sum all the strips, to 
obtain the required surface ABC, x varying from to a. 



Hence 



5 



Jo Jo 



\/a'2—x2 



dx dy 



7r(r 



V a 2 — x* — y l 



Another example is the following 

The centre of a 
sphere, whose radius is 
a, is on the surface of a 
right circular cylinder, 
the radius of whose 
a 



base is 



Find the 




surface of the sphere 
intercepted by the 
cylinder. 

Take for the equa- 
tions of the sphere and 
cylinder, 

and xr + y 2 = ax. 



CPAQ is one fourth 
the required surface. Since this surface is a part of the sphere, the 



358 INTEGRAL CALCULUS 

partial derivatives — , — must be taken from x 2 -f y 2 -f z 2 
ox dy 

giving, as in the preceding example, 



J *S -\i n^ nt& 



-\a 2 — x 2 — y 2 

to be integrated over the region OR A, the projection of CPAQ on 
the plane 17. 

The equation of the curve OR A is x 2 + y 2 = ax. 

Hence ±S = f f^ <***** _ frA ^ 

#=(2ir-4)a 2 . 

Let us now find the surface of the cylinder intercepted by the 
sphere, one fourth of which is CPARO. 

Since this is a part of the cylinder x 2 + y 2 = ax', the partial 
derivatives in (1) must be taken from this equation. But from 

x 2 + y 2 =ax, we find — = oo, — = co- 
ox oy 

The formula (1) is, then, inapplicable in this case. 

It is also evident from the figure that the surface CPARO cannot 
be found from its projection on the plane XY, since this projection 
is the curve ORA. 

The difficulty is removed by projecting on the plane XZ, and 
using, instead of (1), 



We now find from x 2 + y 2 = ax, 

dy a — 2x dy 
dx~ y dz 

Substituting in (2), and simplifying, 



= 0. 



1 o C C adx dz 



2 ^Jax—x 2 



SURFACE, VOLUME, AND MOMENT OF INERTIA 359 



This must be integrated over the region CP'AO, CP'A being the 
projection on XZ of CPA. 

To find the equation of CP'A, we eliminate y from 



giving, 
Hence 



x 2 + y 2 + z 2 = a 2 and x 2 + y 2 = ax, 
z 2 = a 2 —ax. 



\ S ~li i 



v/ q t: 



dx dz 



v 



ax — ar 



EXAMPLES 

1. The axes of two equal right circular cylinders, a being the 
radius of base, intersect at right 
angles : find the surface of one inter- 
cepted by the other. 

Take for the equations of the 
cylinders, 

x 2 -\-z 2 = a 2 , and x 2 -{-7j 2 =a 2 . 

Ans. 8 a 2 . 



2. Find the area of the part of the 
plane 

a o c 

in the first octant, inter- 
cepted by the coordinate 

planes. 




An*- ^V^c-' + c^ + a^. 

3. Find the area of the 
surface of the cylinder 
x 2 -f- y 2 = a 2 , included be- 
tween the plane z=mx and 
the plane IT. Ans . ± ma? . 




ccvJ 



360 



INTEGRAL CALCULUS 



4. Find the area of the surface of the paraboloid of revolution 

y 2 -\-z 2 = 4 ax, intercepted by the 

parabolic cylinder y 2 = ax, and 

the plane x = 3 a. ^q * 

Ans. — ^— . 



5. In the preceding example, 
find the area of the surface of 
the cylinder intercepted by the 
paraboloid of revolution and the 
given plane. 



Ans. (13VT3-1) 



V3 




6. Find the area of that part of the surface 
z 2 -f (x cos a + y sin a) 2 = a? 

which is situated in the first octant. 

The surface is a 
right circular cyl- 
inder, whose axis 
is the line z = 0, 
x cos a+y sin a=0, 
and radius of base a. 

a a 2 
Ans. — . 



sin a cos a 




7. A diameter of 
a sphere whose 
radius is a is the 
axis of a right 
prism with a square 
base, 2 b being the 

side of the square. Find the surface of the sphere intercepted by 
the prism. 

Ans. 8a[2bsm- 1 — — a sin -1 — — -J. 

V Va 2 -6 2 a 2 -b 2 J 



SURFACE, VOLUME, AND MOMENT OF INERTIA 361 

268. To find the Volume of Any Solid bounded by a Surface, whose 
Equation is given between Three Rectangular Coordinates, x, y, z. 

The solid may be supposed to be divided, by planes parallel to the 
coordinate planes, into elementary rectangular parallelopipeds. The 
volume of one of these parallelopipeds is dx dy dz, and the volume of 
the entire solid is r r r 

v= jjj dxd y^ 

the limits of the integration depending upon the equation of the 
bounding surface. 

For example, let us find the volume of one eighth, of the ellipsoid 
whose equation is , 2 2 

a- b 2 <? 



c 












^H 




vif 












m' 








4 










/ 


4. 


< 

P 













^ 


y 




K /'k 'X 



PQ represents one of the elementary parallelopipeds whose volume 
is dx dy dz. 

If we integrate with respect to z, we collect all the elements in the 
column MN\ z varying from zero to MM*\ that is, 



from to z = c x h_''"_ 'L 
V a 1 lr 



362 INTEGRAL CALCULUS 

Integrating next with respect to y, we collect all the columns in 
the slice KLN'H, y varying from zero to KL ; that is, 



Vx 2 
1 . 
a 2 



This value of y is taken from the equation of the curve ALB, 

which is 

^4-^ = 1 
a 2 ^b 2 ' 

Finally, we integrate with respect to x, to collect all the slices in 
the entire solid ABC. Here x varies from zero to OA ; that is, from 
to a. 

The y and x integrations are said to be over the region AOB. 



abc 



a2 I at b2 dxdydz = 

For the entire ellipsoid V= 4 ^?. 



EXAMPLES 

1. Find the volume of one of the wedges cut from the cylinder 
x 2 + y 2 = a 2 by the plane z = x tan a and the plane XY. (See Figure, 
Ex. 3, Art. 267.) 

Ans. 2)1 | dxdydz = 

2. Find the volume of the tetrahedron bounded by the coordinate 
planes and by the plane 

x ' y , z u A abc 

abc o 

3. Find the volume included between the paraboloid of revolution 
y 2 -f z 2 = 4 ax, the parabolic cylinder y 2 — ax, and the plane x=3a. 
(See Figure, Ex. 4, Art. 267.) Ans. (6 tt + 9 V3)a 3 . 



SURFACE, VOLUME, AND MOMENT OF INERTIA 363 

4. Find the volume contained between the paraboloid of revolution 
ar -j- 1/ 2 = «.:. the cylinder x 2 -f y- = 2 ax, and the plane XT. 

2 

5. Find the volume of the cylinder x' 2 + y 2 = ax, intercepted by the 
paraboloid of revolution y 2 + z 2 = 2 ax. fir 2\ 3 

6. The centre of a sphere (radius a) is on the surface of a right 
circular cylinder, the radius of whose base is -. Find the volume of 

the part of the cylinder intercepted by the sphere. (See second Figure, 
^•267.) . 2/ _4\., 



3V 3 

7. Find the volume in the first octant, bounded by the surface 



& 



tt+Pt-l. Am- * 



8. Find the entire volume within the surface 

aj* + y* + «*»<!#. Ans. ^f. 

269. Moment of Inertia of Any Solid. This may be expressed by a 
triple integral. 

Thus, the moment of inertia about OX, m being the mass of a unit 
of volume, is 

I=mJJf(y 2 + z 2 )dxdydz, 

with similar formulae for the moments of inertia about the axes 
OY, OZ. 

EXAMPLES 

1. Find the moment of inertia about OX of the rectangular paral- 
lelepiped bounded by the planes x = a, y = b, z = c, and the co- 
ordinate planes. I . ,, 2 . »< mabc 



364 INTEGRAL CALCULUS 

2. Find the moment of inertia about OZ of the tetrahedron 
bounded by the plane 

a b c 
and by the coordinate planes. Ans. (a 2 -f b 2 ) — — . 

3. Find the moment of inertia about OX of the portion of the 
cylinder x 2 -f y 2 = a 2 included between the planes z = h and z = — h. 

A oj fa 2 . 2 h 2 

Ans. irmarh ( — ■ + -=- 

4. Find the moment of inertia of the preceding cylinder about OZ. 

Ans. TrmaVi. 

5. Find the moment of inertia of a sphere (radius a) about a 
diameter. A 8 -n-ma 5 

6. Find the moment of inertia about OZ of the ellipsoid 



- 2 + 6 - 2 +- 2 - 1 ' ^ s - —15— 



CHAPTER XXXII 

CENTRE OF GRAVITY. PRESSURE OF FLUIDS. 
FORCE OF ATTRACTION 



CENTRE OF GRAVITY 

270. Definition. The centre of gravity of a body is a point so 
situated that the force of gravity acting on the body produces no 
tendency to rotate about an axis passing through the point. 

271. Coordinates of Centre of Gravity. To find the centre of 
gravity. C, of any body, take P as any infinitesimal part of the 
given body, PQ' the line 
of direction of gravity, 
and MN any horizontal 
axis passing through C. 
Let BD be the common 
perpendicular between 
MN and PQ. Take the 
axis of X parallel to BD 
and represent by x and 
x, OL and OL', the x 
coordinates of P and C re- 
spectively. Then the dis- 
tance BD = L'L = x — x. 

The force exerted by gravity on P is proportional to and there- 
fore may be measured by its mass. Denoting its mass by dm, the 
moment of this force about MN would be (x — x)dm; and if dm is 
an infinitesimal in one, two, or three dimensions, the tendency of 




the whole body to rotate about MN is equal to J (as 



x)dm. 



365 



366 INTEGRAL CALCULUS 

Since this must equal zero, 

I (x— x)dm = 0, 

J xdm 



and 



/■ 



a) 



dm. 



Similar formulae may be derived for y and z. 

Note. — -The mass of a unit's volume is called density. If we 
represent the density by p, the differential mass or dm is equal to p 
multiplied by the differential of the arc, area, or volume. 

Ex. 1. Find the centre of gravity of a quarter of the arc of a circle. 

Let the equation of the circle be x 2 + y 2 = a 2 . 

Here dm = p ds. 

Substituting in (1), Art. 271, we have 



x = 



Pxds aC^xia'-x 2 ) ^dx 9 a 



i 

2 wa 



From the symmetry of the figure y = 



2a 




Ex. 2. Find the centre of gravity of the surface bounded by a 
parabola, its axis, and one of its ordinates. 

Let the equation of the parabola be y 2 = ±px, B being (9p, 6p). 
Here dm = p dx dy, and substituting in formula (1), Art. 271. 



x = 



J i x x dx dy 
o Jo 

Jo Jo 

VI p I x 9 -dx 

Jo _ 

p a x 2 dx 

Jo 



27 




V4 



CENTRE OF GRAVITY. PRESSURE OF FLUIDS 307 

Similarly, 



*/ = 



rx** r( g *-£ 



% 



9p 
4 



Ex. 3. Find the centre of gravity of a circular disk of radius 
a, whose density varies directly as the distance from the centre, and\ 
from which a circle described 
upon a radius as a diameter 
has been cut. 

Let the equation of the 
large circle be r = a ; and 
the equation of the small 
circle be r = — a cos 0. 

The disk is s}*mmetrical 
with respect to OX, hence 

»=o. 

Here 
dm = pr dd dr = kt 2 dd dr, 

(if p = Kr). 

Also x = OM= r cos 0. 




Therefore x 



Jo «yo 



r 3 cos dd dr 



%Jtt_ %J — I 



?* 3 cos dO dr 



cos 9 



dd dr 



x = 



2 | r 2 cos d dd dr +11 r 2 cos 

c/o %/0 «/7r J —acos 9 

T f f cos dd 4- f * (cos $ - cos 5 0) dB 1 



[P*+£ 



(1 + cos 3 0) c76> 



] 



5(3tt-2) 



= 0.1016 a. 



368 



INTEGRAL CALCULUS 



Ex. 4. Find the centre of gravity of a cone of revolution, the 
radius of the base being 2 and the altitude 6. 
The equation of OB is y = ±x. 
Here dm = piry 2 dx, and substituting in (1), Art. 271, 



| xy- dx 

.Jo 



y 2 dx 



The cone is symmetrical with 
respect to OX, hence y = 0. 

Note. — On comparing the 
formulae for the centre of grav- 
ity of arc, area, and volume, 



Y 


X 
^^ * .1 


i \ 




3 




o^- cl 


1 1 






X 



/xds J xdA I ; 

4V — j *Aj — ■ y *As » — 

f ds f dA 



far 



we notice that, in each case, the element of the numerator integral 
is x times the element of the denominator integral. 



5. Find the centre of gravity of the arc of the hypocycloid 

2 2. 2 _ _ 2 

(Art. 125) x* + y* =a f in the first quadrant. Ans. x = y — -a. 

o 

6. Find the centre of gravity of the arc of the cycloid 



-iV 



x = a vers" 1 ± — V2 ay — y 2 . 



Ans. x = -n-a, y = -a. 



7. Find the centre of gravity of a straight rod of length a, the 
density of which varies as the third power of the distance of each 



point from the end. 



Ans. x = -a. 
5 



8. Find the centre of gravity of the surface of a hemisphere 
when the density at each point of the surface varies as its perpen- 
dicular distance from the base of the hemisphere. * - _ 2_a 



CEXTRE OF GRAVITY. PRESSURE OF FLUIDS 369 
9. Find the centre of gravity of a semiellipse. Ans. z = —-. 

3 7T 

10. Find the centre of gravity of the area between the cissoid 

y- = — and its asvinptote. Ans. x = - a. 

2 a — x 3 

11. Find the centre of gravity of the area bounded by the parab- 
ola y- = 8 x, the line y -f- x — 6 = 0, and the axis of X. 

Ans. a = 2.48; y=lA. 

12. Find the centre of gravity of one loop of the curve r = a sin 2 $. 

A - 128 a _ 128 a 
Ans. x — ; y = 

105 7r' J lOOTT 

13. Find the centre of gravity of the upper half of the cardioid 

' = «(1-«*<D- Ans , aj = _5 a . 

6 ' 

_ 16 a rrr 

y = ——=.57a. 

14. Find the centre of gravity of one loop of the lemniscate 
rW ° 0s2( '' Ans x- W *a-55a 

^cxllb. X — — — LI — .OO LI. 

8 

3 

15. Find the centre of gravity of a hemisphere. Ans. x = -a. 

8 

3 

16. Find the centre of gravity of a hemispheroid. Ans. '% = - a. 

8 

17. A cone of height h is scooped out of a cylinder of the same 
height and base. Find the distance of the centre of gravity of the 
remainder from the vertex. , 3 , 

8 

272. Theorems of Pappus. 

Theorem I. If a plane area be revolved about an axis in its plane 
and not crossing the area, the volume of the solid generated is equal 
to the product of the area and the length of the path described by 
the centre of gravity of the area. 



370 INTEGRAL CALCULUS 

Theorem II. If the arc of a curve be revolved about an axis in 
its plane and not Crossing the arc, the area of the surface generated is 
equal to the product of the length of the arc and the path described 
by the centre of gravity of the arc. 

273. Proof of the Theorems. Let the area be in the plane XY 
and let it revolve about the axis of X. Then by (1), Art. 271, we 
have 



y = 



{ (ydxdy 
I I dxcly 



Or yj j dx dy=j \y dx dy. 

Then 2 Try f Cdx dy = C (*2 Try dx dy (1) 

But the right-hand member of equation (1) is the volume described 
by revolving the area through the angle 2 it, 2 tig is the length of the 

path described by the centre of gravity, and j I dxdy is the plane 
area. 

The first theorem is thus seen to be true, and the second can be 
proved true in a similar manner. 

EXAMPLES 

1. Find the volume and surface generated by revolving a rec- 
tangle with dimensions a and b about an axis c units from the centre 
of the rectangle. Am 2 ^^ and 4 ^ + &)c 

2. Eind the volume and surface generated by revolving an equi- 
lateral triangle each side a units in length, about an axis c units 
from the centre of the triangle. 2 r= 

Am. ' 7rac ^ 6 and 6 fl-ac. 
2 



CENTRE OF GRAVITY. PRESSURE OF FLUIDS 371 

3. Find the volume and surface generated by revolving a circle 
of radius a about an axis c units from the centre of the circle. 

Ans. 2 irarb and 4 -r^ab 



4. Find the volume generated by revolving an ellipse, semiaxes 
a and b, about an axis c units from the centre of the ellipse. 

Ans. 2 7r 2 abc. 



PRESSURE OF LIQUIDS 

274. The pressure of a liquid on any given horizontal surface is 
equal to the weight of a column of the liquid whose base is the 
given surface and whose height is equal to the distance of this sur- 
face below the surface of the liquid. 

The pressure on any vertical surface varies as the depth, and the 
method of determining it is illustrated by the following examples. 

Ex. 1. Suppose it is required to find the pressure on the 
rectangular board OABC, the edge OC being at the surface of the 
water. 

Let BC= a, and AB — b. C Y 

Suppose the rectangle divided 
into horizontal strips one of which 
is HK. 

Let OH=x, then the width of 
the strip is dx. 

If the pressure on this strip 
were uniform throughout and the 
same as it is at the top of the strip, 
the pressure on the strip would be 
wbx dx, where iv is the weight of 
a cubic unit of the water. And 
the entire pressure on the board is 
expression. 



H 



X 




a 




K 




B 




b 



evidently the integral of this 



That is, Entire pressure 



-r- 



bx dx = 



a-biv 



372 



INTEGRAL CALCULUS 



Ex. 2. Find the pressure on that part of the board in Exam- 
ple 1, which is below tile diagonal. 

In this case the area of UK is y dx, and the entire pressure on the 
triangular board is 



f 



But 



wyx dx. 
b 



hence entire pressure 

bw C a o 7 
= — I x~dx = 

a Jo 



bwa? 




Ex. 3. One face of a box immersed in water is in the form of 
a square, the diagonals being 8 feet in length. The centre of the 
square is 6 feet below the surface of the water, and one diagonal is 
vertical. Eind the pressure 
on the square face. ~ YV 

Let SW be the surface 
of the water. Taking the 
axes as in the figure, the 
equations of AB and BO 
are y=4-f x, and ?/=4— x, 
respectively. 

Then, if P represents the 
entire pressure on the 
board, 

P=2iv( C (6-{-x)dydx 
= 512 tv = 15872 lbs. 




Ex. 4. Eind the pressure on a sphere 6 feet in diameter, im- 
mersed in water, the centre of the sphere being 10 feet below the 
surface of the water. 



CENTRE OF GRAVITY. PRESSURE OF FLUIDS 373 



Let SWbe the surface of 
the water. 

Take the axes as in the 
figure, and let the elemen- 
tary surface be a zone. 
The area of a zone at a 
distance x from the cen- 
tre of the sphere is 2 -y ds. 
The pressure on the zone 
is 2 -"vmIO + a?)efc. 

Then, if P represents 
the entire pressure on the 
sphere. 



W 




m L y 



(10 + x)ds. 



But 

Hence 



y = V9 — x 2 , and ds = - civ. 

y 



P=far«fl C (10 + x)dx, 



= 360 ttv: = 22320tt lbs. 

5. A rectangular flood gate whose upper edge is in the surface of 
the water, is divided into three parts by two lines from the middle 
of lower edge to the extremities of upper edge. Show that the parts 
sustain equal pressures. 



6. A rectangular flood gate 10 feet broad and 6 feet deep has its 
upper edge in the surface of the water. How far must it be sunk to 
double the pressure ? Ans. 3 ft. 

7. A board in the form of a parabolic segment by a chord perpen- 
dicular to th< j axis is immersed in water. The vertex is at the sur- 
far-p and the axis vertical. It is 20 feet deep and 12 feet broad. 
Find the pressure in tons. Ans. 59.52. 



374 INTEGRAL CALCULUS 

8. How far must the board in Ex. 5 be sunk to double the 
pressure ? Ans. 12 ft. 

9. Suppose the position of the parabolic board in Ex. 5 reversed, 
the chord being in the surface ; what is the pressure ? 

Ans. 39.38 tons. 

10. How far must the board in Ex. 7 be sunk to double the 
pressure ? Ans. 8 ft. 

11. A trough 2 feet deep and 2 feet broad at the top has semi- 
elliptical ends. If it is full of water, find the pressure on one end. 

Ans. 165^ lbs. 

12. One end of an unfinished water main 2 feet in diameter is 
closed by a temporary bulkhead and the water is let in from the 
reservoir. Find the pressure on the bulkhead if its centre is 30 feet 
below the surface of the water in the reservoir. Ans. 18607r lbs. 

13. A water tank is in the form of a hemisphere 24 feet in diame- 
ter surmounted by a cylinder of the same diameter and 10 feet high. 
Find the total pressure on the surface of the tank when the tank is 
filled to within 2 feet of the top. Ans. 148. 8?r tons. 

14. A cylindrical vessel, whose depth is 12 inches and base a 
circle of 20 inches diameter, is filled with equal parts of water and 
oil. Assuming the oil to be half as heavy as the water, show that 
the pressure on the base equals the lateral pressure. 

275. Centre of Pressure. Since the pressure of a liquid on a verti- 
cal surface varies as the depth, there exists a horizontal line about 
which the statical moment of the entire pressure on the surface is 
zero. Such a line passes through the centre of pressure and the 
abscissa of this point may be found by the method used in the follow- 
ing example. 

Ex. 1. Find the abscissa of the centre of liquid pressure on 
a vertical surface bounded by the curve y = f(x), the axis of X and 
the two ordinates y Q and y v Given that the origin is at a distance 



CENTRE OF GRAVITY. PRESSURE OF FLUIDS 375 



h below the surface of the liquid, the axis of X vertical, and the 
-weight of a cubic unit of liquid is w. 

Let P^JPJIQ be the surface bounded by the curve y = f(x), the 
axis of X, and the two ordinates ?/ = QP and y x — HP V Divide 
the surface into horizontal strips of width dx, one of which is HK. 
Let OH = x. Let MN pass through the centre of liquid pressure, 
and 031— x. 

Then the pressure on the 
strip HK is wy (h + x) dx, 
and the moment of this 
pressure about MN is 
wy(h + x)(x —I-)dx. 

Therefore, the moment 
of the entire pressure is 
the integral of this ex- 
pression between the ab- 
scissas of P and P l} that 
is, between x and x 1 . 
But this must equal zero, 
therefore 



£ 



Or 



wy (J t + x) (x - x) dx = 0. 

1 xy (h -f a;) dx 

x=^ x . 

Jy(h + x)dac 











— W 






|% 

















Y 
















! Pn 




Q 




X 


\x 


\ M 


M 


l i«» 


H 





IS. 


R 










I J 1 


X 













2. Find the centre of pressure of the water on the parabolic 
board given in Ex. 7, Art. 274. Ans. 14f in. below vertex. 

3. Find the centre of pressure of the water on the bulkhead given 
in Ex. 12, Art. 274. Ans. x = T ^ ft. 

4. A rectangular flood gate a feet deep and b feet broad, with its 
upper edge at the surface, is to be braced along a horizontal line. 
How far down must the brace be put that the gate may not tend to 
turn about it ? Ans. | a ft. 



376 INTEGRAL CALCULUS 

5. One end of a cylindrical aqueduct 6 feet in diameter which is 
half full of water is closed by a water-tight bulkhead held in place 
by a brace. How far below the centre of the bulkhead should the 
brace be put ? What pressure must it be able to withstand ? 

Ans. x = T 9 F 7T ft. ; P = 1116 lbs. 

6. A water pipe passes through a masonry dam, enters a reservoir, 
and is closed by a cast-iron circular valve which is hinged at the 
top. The diameter of the valve is 3 feet, and the depth of its centre 
below the water level in the reservoir is 12 feet. Find the pressure 
on the valve, and the distance of the centre of pressure below the 
hinge. Ans. P = 1674 tt lbs. and ff ft. 

7. Water is flowing along a ditch of rectangular section 4 feet 
deep and 1 foot wide. The water is stopped by a board fitting the 
ditch and held vertical by two bars crossing the ditch horizontally, 
one at the bottom and the other one foot from the bottom of the 
ditch. How high must the water rise to force a passage by upset- 
ting the board ? Ans. To within 1 ft. of top of ditch. 

ATTRACTION AT A POINT 

276. A particle of mass m is situated at a perpendicular distance 
c from one end of a thin, straight, homogeneous wire of mass M and 
length I. Required to find the attraction on the particle due to the 
wire. 

Let be the particle and AB the wire. Let X and Y be the 
components of the attraction along the axes of X and ^respectively. 

Divide AB into elements of length dy 
and let PQ be one of these elements. 

M M 

The mass of PQ is — dy, since — is 

l L 

the mass of a unit's length. 

If the mass of PQ were concen- 
trated at P, the attraction at due 
to PQ is, according to Newton's Law 
of Attraction, KmJ(f(% 

l(c 2 + f)' 




CENTRE OF GRAVITY. PRESSURE OF FLUIDS 377 

and the components along OX and OF are 

«"»* d J* cos 6, and *™ Md l sin fl, 

respectively. 

Substituting for cos and sin their values we have 

^ KinMc C l d\j KinM KinM . a 
X = I •- = — — — -_ = sm 6. 

I JoQS + y*)? cW^+T 2 Cl 

TT -_ KinM C 1 y dy _ KinM f^ c ~| 

^° (c' + yrf cl L V?+T 2 J 



I ^° (c 2 + y 2 y cl 

_ KinM 



Vc 2 + 
(1 — cos 



Denoting by It the total attraction of the wire on the particle, 
•K= VI 2 + Y 2 = —£- V2(l - cos 0) 



2 KmM . 1 „ 

sm-0. 

cl 2 



The line of attraction evidently makes with OA an angle whose 
tangent is 

J^l-cos^lg 
X sin 2 

The resultant attraction, therefore, bisects the angle 6. 

Xote. — If we take as our unit of force the force of attraction be- 
tween two unit masses concentrated at points which are at unit's 
distance apart, k becomes unity. 



378 INTEGRAL CALCULUS 



EXAMPLES 



1. Find the attraction perpendicular to the wire in Example 1 



when the particle is at a distance - above 0. 

o 

Ans. SiM 



21 . I 



>] 



c [_V9 c- + <U 2 V9c- + 



2. Find the attraction of a thin, straight, homogeneous wire of 
length I and mass M upon a particle or mass m which is situated at 
a distance c from one end of the wire and in its line of direction. 

KinM 



Ans. 



c(c + l). 



3. Find the attraction of a homogeneous circular disk of radius a 
upon a particle of mass m in its axis and at a distance c from the 
disk. 

where p is the density of the disk. 



Ans. 2 Kirmpl 1 — 






Vc 2 + a 2 J 



4. Find the attraction due to a homogeneous right circular cylinder 
of length 2 I and radius a upon a mass m in the axis produced of the 
cylinder and distant c from one end. 



Ans. 2 TTKinp [2 ?+Va 2 + c 2 - Va 2 + (c + 2 J) 2 ]. 



CHAPTER XXXIII 

INTEGRALS FOR REFERENCE 

277. We give for reference a list of some of the integrals of the 
preceding chapters. 

/ r n+l 
x" dx = 
ra-fl 



2. I — = log 05. 

J X 



o C dx 

3. I-— — =-tan 

J xr -f- a." a a 



-r cr a a 

- r dx 1 , # — a 

4. I — = — log 

J x 2 — a 2 2 a x-\-a 

EXPONENTIAL INTEGRALS 



5. Ca x dx = 



a x 
log a 



6. 



j e x dx = e x . 



TRIGONOMETRIC INTEGRALS 

7. I sin x dx = — cos x. 



j sin x dx = 
8. I cos £ dx = sin x. 



379 



380 INTEGRAL CALCULUS 

9. I tan x dx = log sec x. 

10. I cot x dx = log sin x. 

11. a sec x dx = log (sec x + tan x) 

= logtan(|+|). 

12. I cosec x dx = log (cosec x — cot x) 

= log tan - • 

13. j sec 2 x dx = tan x. 

14. j cosec 2 cc dx = — cot a?. 

15. I sec x tan x dx = sec x. 

16. j cosec a? cot x dx = — cosec #. 

17. j sin 2 xd£ = - --sin2a\ 



2 4 



18. I cos 2 a? dx = - -f - sin 2 #. 

2 4 



/< 



INTEGRALS CONTAINING Vcr-a; 2 (CHAP. XXV. AND ART. 227) 

- n C dx . _!# 

19. I — — sin -• 

J Va 2 — sc 2 a 

20. f _^L_ = _ 2 V^^t? + - sin- 1 - • 
J Va 2 -^ ^ 2 2 a 



INTEGRALS FOR REFERENCE 381 

21. 



C dx 1, x 

I = - log = 

^ sca a 2 — or a a + a ((- 



ar 



22 



23 



r da? __ _ a a 2 -.r 
•^ a* 2 a a 1 ' — x* (r>r 

J . r \ cr _ x' 2 2 a-.r 2 a 3 °° a + ya 2 -^ 2 ' 

24. (a ^^ da? = | Vtf^a? + - sin- 1 - - 

25. f.r\'^7 da? = % (2 ar 2 - a 2 )Va 2 -r + - sin" 1 - . 
J 8 8 a 

26. f — = (Art. 227.) 

•/ (a*- a? 2 )* « 2 a cr-x 2 

27. f (a 2 - :r) I (to = 2 (5 a 2 - 2 x 2 ) Va^=^ + — sin" 1 ? . 
*/ b 8 a 



INTEGRALS CONTAINING Va? 2 + a 2 (CHAP. XXV. AND ART. 227) 

28. f ** = log(a?4-VS*"+^). 

^ a .r + a 2 

29. (' jrVj? =^A^T^-giog(.x-4-A^T^). 
c/ A ■ _ ,,- 2 J 



Jdaj 1 , .x* 1 t v a? 2 4- a 2 — a 

— = log = - 1 — 

x\ .- - a 2 a a -f Var' -f a 2 ct 



31 f da; = _ Vx 2 + a s 

J 2 a?2 



382 INTEGRAL CALCULUS 



34 



35. 



J or 5 Vx 2 4- a 2 2aV 2 a 3 a 

33. fV^T^tto = | V^T^ + f log (a + -y/iF+a*). 

. fa^Va? 2 + a 2 dx = x (2x 2 + a^Va^ + a 2 - - log (jc + VV + a 2 ). 
Jo 8 

/dec __ % 

(x 2 +a 2 )* aV^ + a 2 

36. f(x 2 + a 2 ) 1 da> = ^ (2 x 2 + 5 a 2 ) V^+"^ + ~ log (»+ Va?+a*). 
%/ 8 8 

INTEGRALS CONTAINING Va 2 - a 2 (CHAP. XXV. AND ART. 227) 

37. f da; = log(a? + V^a»). 
J Vec 2 — a 2 



38. f / ^_g_, = ? V* 2 - a 2 + £ log (« + V?=^) 



oa {* dx 1 , x 

39. I -———- = - sec -1 - 

•^ xV# 2 — a 2 a a 



40 



/: 



c?» V x 2 — a 2 



x 2 -\/x 2 — a 2 a ' x 



dec V# 2 — or . 1 _i a 

^ ==== = 1 sec - • 

x^x 2 ~a 2 2aV 2 a 3 a 



INTEGRALS FOR REFERENCE 383 

42. (\ .r-crV.r = | a .r - «- - | log (x + Vx 2 - a 2 ). 

43. faj 2 \ Z^T? <fe = 5 (2 or - a 2 ) V^ 2 ^ 2 - ^ log (« + Vtf 2 -a*). 

•J 8 b 

44. f dg = * • 

•* a-- — cr)^ «Vr — a 2 

45. f^r-a 2 )^7.r = ^(2^-5a 2 )V^^ + ^log(a;+V^ 4rr ^). 
»y 8 8 



INTEGRALS CONTAINING V2 a.r - x 2 

46. I — = vers ! -• 

J V2 cub 



sr 



a 



47. I — = — V 2 a.? — a^+ a vers ' - • 

•^ V2oa?— x 2 a 

a*£ V2 «.i* — x 2 



48 






V2 aa — x' 1 



49 . ( v 2 ax — a; 2 tfte = - — -• V2 ax — x 2 + ^- vers -1 - • 
J 2 2 a 

50. f«v?^=7* = - 3a ' + <f- 2 * VU^rf + £ vers-*. 
J 6 2 a 

51. f>^^-^^= v aax-^ + avers- 1 *. 

J u; a 



384 INTEGRAL CALCULUS 

s 



. W2ax — x 2 dx _ _ (2 ax — x 2 ) 2 
x 3 3 ax 3 



/» dx x — a 

53. f 



54. f 



(2 ax — x-y a 2 V2 ax — x 2 
xdx x 



(2ax — x 2 y aV2ax — 



INTEGRALS CONTAINING ±ax 2 +bx + c 

55 C_J* = 2 tan-i 2^ + 6 , 

' J aar + 6a; + c V4ac-6 2 V4ac-6 2 



1 , 2aa; + 6- V6 2 -4ac 

56. or = , ~-\og - , 

V6 2 — 4ac 2aa; + 6 + V6 2 — 4 etc 

57. f , 9 , = = — = log (2 aa; + 6 + 2vWaa; 2 + 6a; + c). 
J V aar + 6a; + c V a . 

58. 



I- I -y/ax 2 -\-bx + cdx = — Vax 2 + 6a' 

^ 4a 



6 2 — 4 ac 



8 a* 

1 f ^ = 1 

•^ V — aa^ + 6a; + c Va 



log (2 aa; + 6 + 2VaV aa; 2 + 6a; + c). 



_i 2 aaj — 6 
59. I — = — = sin x — — 



V 6 2 + -i ac 



60. 



j V — aa; 2 -\-bx + cdx 



2 aa; — 6 / *——, — ; — , 6 2 + 4 ac . _, 2 aa; — 6 

— V — ax 2 -{- bx -\- c -{ — sin * — z^=. 

4 a 8a | V6 2 + 4ac 



INTEGRALS FOR REFERENCE 385 

OTHER INTEGRALS 



61. CJ^t^cU 
J \6 4-a- 



+ 



a {a + .i-) (b + x) + (a - 6) log ( Va + a? + V& + x). 



62. 



f JjL^cfa = V(a - a)(& + 0) + (a + 6) sin" 1 J^|- 



INDEX 



Acceleration 18 

Angles, between two curves . . 174, 183 
with coordinate planes . . 133 

Arc, derivative of 186, 187 

Area, any surface 355 

derivative of 241 

of curve . . 242, 306, 320, 325, 349 
surface of revolution . 337, 353 

Asymptotes 179 

Attraction at a point 375 

Cauchy's test for convergence . . 82 

Centre, of curvature 200 

of gravity 365 

of pressure 373 

Change of variable . . .57, 58, 148, 263, 

299, 304, 317 

Circle, of curvature .... 195, 200 

osculating 209 

Comparison test for convergence . . 80 
Computation, by logarithms ... 94 

of n 96 

Constant, definition of 1 

derivative of 26 

notation of 1 

of integration . . . . 224, 315 
Contact, order of ... . 206, 207, 209 
Convergence, absolute and condi- 
tional 78 

interval of .... 86 

tests for 79 

Curvature, centre of 200 

circle of ... . 195, 200 

direction of 189 

radius of . . . 193, 196, 197 

uniform 193 

variable 194 

Curves, angle of intersection of . . 197 
area of . . 242, 306, 320, 325, 349 
continuous and discontinu- 
ous 22 



Curves, direction of . . . 16, 174, 182 
for reference, higher plane . 162 

length of 327, 330 

osculating 208 

Definite integrals 307 

as a sum . . . 319 
definition of . . 310 
double .... 343 
sign of .... 314 

Differential coefficient 12 

Differentials, definition of . . . 68, 70 

formulae 71 

Differentiation, algebraic formulae 26, 29 
definition of . . . 26 
inverse trigonomet- 
ric formulae . . 51 
logarithmic and ex- 
ponential formu- 
lae 39 

order of .... 137 

partial 130 

successive .... 61 
trigonometric formu- 
lae 45 

Derivative, definition 11 

general expression of . . 12 
illustrations of ... . 13 
meanings of . . . 16, 17, 18 

of an arc 186, 187 

of area .... 241,336 
of function of a func- 
tion 58 

partial 130 

partial, of higher order 

136, 137 
dv 



relation between — 

ax 

and *L 

total ...... 



57 
140 



386 



INDEX 



387 



TACK 

Element, of area 320 

of an integral 319 

Envelopes, definition of . . . 214, 215 

equation of 21S 

of normals 217 

Equation, of envelopes 215 

ofevolute 201 

of normal 133 

of tangent 133 

parametric 324 

Evolute 201 

an envelope 217 

equation of 201 

properties of ... . 202, 204 

Function, algebraic 2 

continuous 22, 315 

definition 1 

discontinuous 22 

expansion of 88 

implicit) differentiation of 

75, 144 

increasing and decreasing 21 

inverse trigonometric . . 51 

logarithmic 39 

of two or more variables . 15G 

transcendental .... 2 

trigonometric ..... 45 

Gravity, centre of 365 

Higher plane curves 162 

Huyghens's approximate length of 
arc 93 

Increment 11 

Indefinite integral 309 

Indeterminant forms 106 

Inertia, moment of 340, 363 

Infinite, limits 315 

variables 315 

Infinitesimals, order of 73 

Inflexion 190 

Integration and derivative of integral 270 
as a summation .... 306 
between limits .... 309 
by algebraic substitution 299 

by parts 279 

by substitution . . 218-22*; 
constant of . . . 224, 310 
containing 
Va*— x*, . - . 296 



PAGE 
P 

Integration, containing (ax+b)? . . 263 
containing 

p r 

(ax+b)v, (ax + b)* . 264 
containing 

V± x 2 + ax + b ... 266 

definite 307 

definition of 223 

double . 343, 347, 349, 350 

evaluation of .... 307 

for reference .... 378 

fundamental 225 

indefinite 309 

of see" x dx, cosec" x dx . 274 

of sin" x dx, cos' 1 x dx . 270 
of sin m x cos n x dx 

271, 276, 291, 293 

of tan" x dx, cot" x dx . 273 
of tan" x sec" x dx, 

cot" x cot" x dx 274, 291, 293 
of e^ sin nx dx, 

e ax cos nx dx 283 

of x m (a + bx n )Pdx . . . 284 

of f(x 2 )xdx 295 

of rational fractions . . 249 

proofs of formulae . . . 227 

successive 343 

triple . ... 344, 361, 363 

Intercepts of tangent 173 

Involute 201 

properties of ... . 202, 204 

Leibnitz's theorem 65 

Length of curves .... 93, 327, 330 

Limit, change of 317 

definition of 8 

infinite 315 

Napierian base 10 

notation of 8 

relation of arc to chord . . 9 

variable 343, 348, 351 

Liquids, pressure of 370 

Logarithmic functions .... 2, 39 

Logarithms, computation by . . . 94 

Napierian 41 

Maclaurin's theorem 89 

Maxima and minima .... 114, 155 
Moment of inertia 346, 363 



388 



INDEX 



PAGE 

Napierian base . . „ 8 

Normals , .... 133 

Order, of contact ....... 206 

of differentiation 137 

of integration 343 

Osculating curves 208 

order of contact of . . 209 
Osculating circle, coordinates of centre 

209 
radius of 209 

Pappus, theorems of 369 

Parameter 214, 324 

Parametric equations ..... 324 

Power series 85 

Pressure, centre of 373 

of liquids 370 

Rates 18 

Reduction formulae 284, 291 

Remainder, Taylor's theorem ... 105 

Series, computation by 94 

convergence of power ... 85 
convergent and divergent . . 78 
of positive and negative 

terms 78 

power 85 

Slope of a curve 16 

of a line 16 

of a plane 133 

Subtangent 173, 183 

Subnormal . 173. 183 



PAGE 

Surfaces, area of any 355 

Surfaces of revolution, areas of 337, 353 

derivative of 
area of . . 336 

volumes of . 333 

Tangent 70 

intercept of 173 

Tangent planes 133 

Taylor's theorem .... 97, 103, 145 

Theorem, Leibnitz's 65 

Maclaurin's 89 

mean value . . . . 84, 86 

Pappus's 369 

Rolle's 83 

Taylor's ... 97, 103, 145 

Transformation 152, 153 

Trigonometric functions ... 2, 45 

Uniform curvature . „ . . . . 193 
Unit of force . . . „ ... . 376 

* 

Variable, change of . . . 57, 58, 148, 

263, 299,304, 317 

curvature ...... 194 

definition of 1 

dependent 2 

independent 2 

notation of 1 

Velocity 17, 18 

Volumes, any solid 361 

by area of section . . . 340 
surfaces of revolution 333, 353 



